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442 lines
18 KiB
Markdown
442 lines
18 KiB
Markdown
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---
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comments: true
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---
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# 12.2 Divide and conquer search strategy
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We have learned that search algorithms fall into two main categories.
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- **Brute-force search**: It is implemented by traversing the data structure, with a time complexity of $O(n)$.
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- **Adaptive search**: It utilizes a unique data organization form or prior information, and its time complexity can reach $O(\log n)$ or even $O(1)$.
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In fact, **search algorithms with a time complexity of $O(\log n)$ are usually based on the divide-and-conquer strategy**, such as binary search and trees.
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- Each step of binary search divides the problem (searching for a target element in an array) into a smaller problem (searching for the target element in half of the array), continuing until the array is empty or the target element is found.
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- Trees represent the divide-and-conquer idea, where in data structures like binary search trees, AVL trees, and heaps, the time complexity of various operations is $O(\log n)$.
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The divide-and-conquer strategy of binary search is as follows.
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- **The problem can be divided**: Binary search recursively divides the original problem (searching in an array) into subproblems (searching in half of the array), achieved by comparing the middle element with the target element.
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- **Subproblems are independent**: In binary search, each round handles one subproblem, unaffected by other subproblems.
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- **The solutions of subproblems do not need to be merged**: Binary search aims to find a specific element, so there is no need to merge the solutions of subproblems. When a subproblem is solved, the original problem is also solved.
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Divide-and-conquer can enhance search efficiency because brute-force search can only eliminate one option per round, **whereas divide-and-conquer can eliminate half of the options**.
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### 1. Implementing binary search based on divide-and-conquer
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In previous chapters, binary search was implemented based on iteration. Now, we implement it based on divide-and-conquer (recursion).
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!!! question
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Given an ordered array `nums` of length $n$, where all elements are unique, please find the element `target`.
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From a divide-and-conquer perspective, we denote the subproblem corresponding to the search interval $[i, j]$ as $f(i, j)$.
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Starting from the original problem $f(0, n-1)$, perform the binary search through the following steps.
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1. Calculate the midpoint $m$ of the search interval $[i, j]$, and use it to eliminate half of the search interval.
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2. Recursively solve the subproblem reduced by half in size, which could be $f(i, m-1)$ or $f(m+1, j)$.
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3. Repeat steps `1.` and `2.`, until `target` is found or the interval is empty and returns.
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The diagram below shows the divide-and-conquer process of binary search for element $6$ in an array.
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{ class="animation-figure" }
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<p align="center"> Figure 12-4 The divide-and-conquer process of binary search </p>
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In the implementation code, we declare a recursive function `dfs()` to solve the problem $f(i, j)$:
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=== "Python"
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```python title="binary_search_recur.py"
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def dfs(nums: list[int], target: int, i: int, j: int) -> int:
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"""二分查找：问题 f(i, j)"""
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# 若区间为空，代表无目标元素，则返回 -1
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if i > j:
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return -1
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# 计算中点索引 m
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m = (i + j) // 2
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if nums[m] < target:
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# 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j)
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elif nums[m] > target:
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# 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1)
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else:
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# 找到目标元素，返回其索引
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return m
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def binary_search(nums: list[int], target: int) -> int:
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"""二分查找"""
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n = len(nums)
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# 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1)
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```
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=== "C++"
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```cpp title="binary_search_recur.cpp"
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/* 二分查找：问题 f(i, j) */
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int dfs(vector<int> &nums, int target, int i, int j) {
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// 若区间为空，代表无目标元素，则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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int m = (i + j) / 2;
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1);
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} else {
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// 找到目标元素，返回其索引
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return m;
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}
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}
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/* 二分查找 */
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int binarySearch(vector<int> &nums, int target) {
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int n = nums.size();
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// 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1);
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}
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```
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=== "Java"
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```java title="binary_search_recur.java"
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/* 二分查找：问题 f(i, j) */
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int dfs(int[] nums, int target, int i, int j) {
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// 若区间为空，代表无目标元素，则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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int m = (i + j) / 2;
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1);
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} else {
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// 找到目标元素，返回其索引
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return m;
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}
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}
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/* 二分查找 */
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int binarySearch(int[] nums, int target) {
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int n = nums.length;
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// 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1);
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}
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```
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=== "C#"
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```csharp title="binary_search_recur.cs"
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/* 二分查找：问题 f(i, j) */
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int DFS(int[] nums, int target, int i, int j) {
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// 若区间为空，代表无目标元素，则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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int m = (i + j) / 2;
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return DFS(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return DFS(nums, target, i, m - 1);
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} else {
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// 找到目标元素，返回其索引
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return m;
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}
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}
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/* 二分查找 */
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int BinarySearch(int[] nums, int target) {
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int n = nums.Length;
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// 求解问题 f(0, n-1)
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return DFS(nums, target, 0, n - 1);
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}
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```
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=== "Go"
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```go title="binary_search_recur.go"
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/* 二分查找：问题 f(i, j) */
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func dfs(nums []int, target, i, j int) int {
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// 如果区间为空，代表没有目标元素，则返回 -1
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if i > j {
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return -1
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}
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// 计算索引中点
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m := i + ((j - i) >> 1)
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//判断中点与目标元素大小
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if nums[m] < target {
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// 小于则递归右半数组
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m+1, j)
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} else if nums[m] > target {
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// 小于则递归左半数组
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m-1)
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} else {
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// 找到目标元素，返回其索引
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return m
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}
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}
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/* 二分查找 */
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func binarySearch(nums []int, target int) int {
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n := len(nums)
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return dfs(nums, target, 0, n-1)
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}
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```
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=== "Swift"
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```swift title="binary_search_recur.swift"
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/* 二分查找：问题 f(i, j) */
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func dfs(nums: [Int], target: Int, i: Int, j: Int) -> Int {
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// 若区间为空，代表无目标元素，则返回 -1
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if i > j {
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return -1
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}
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// 计算中点索引 m
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let m = (i + j) / 2
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if nums[m] < target {
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// 递归子问题 f(m+1, j)
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return dfs(nums: nums, target: target, i: m + 1, j: j)
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} else if nums[m] > target {
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// 递归子问题 f(i, m-1)
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return dfs(nums: nums, target: target, i: i, j: m - 1)
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} else {
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// 找到目标元素，返回其索引
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return m
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}
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}
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/* 二分查找 */
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func binarySearch(nums: [Int], target: Int) -> Int {
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// 求解问题 f(0, n-1)
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dfs(nums: nums, target: target, i: nums.startIndex, j: nums.endIndex - 1)
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}
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```
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=== "JS"
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```javascript title="binary_search_recur.js"
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/* 二分查找：问题 f(i, j) */
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function dfs(nums, target, i, j) {
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// 若区间为空，代表无目标元素，则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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const m = i + ((j - i) >> 1);
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1);
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} else {
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// 找到目标元素，返回其索引
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return m;
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}
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}
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/* 二分查找 */
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function binarySearch(nums, target) {
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const n = nums.length;
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// 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1);
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}
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```
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=== "TS"
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```typescript title="binary_search_recur.ts"
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/* 二分查找：问题 f(i, j) */
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function dfs(nums: number[], target: number, i: number, j: number): number {
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// 若区间为空，代表无目标元素，则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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const m = i + ((j - i) >> 1);
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1);
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} else {
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// 找到目标元素，返回其索引
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return m;
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}
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}
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/* 二分查找 */
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function binarySearch(nums: number[], target: number): number {
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const n = nums.length;
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// 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1);
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}
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```
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=== "Dart"
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```dart title="binary_search_recur.dart"
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/* 二分查找：问题 f(i, j) */
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int dfs(List<int> nums, int target, int i, int j) {
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// 若区间为空，代表无目标元素，则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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int m = (i + j) ~/ 2;
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1);
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} else {
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// 找到目标元素，返回其索引
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return m;
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}
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}
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/* 二分查找 */
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int binarySearch(List<int> nums, int target) {
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int n = nums.length;
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// 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1);
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}
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```
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=== "Rust"
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```rust title="binary_search_recur.rs"
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/* 二分查找：问题 f(i, j) */
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fn dfs(nums: &[i32], target: i32, i: i32, j: i32) -> i32 {
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// 若区间为空，代表无目标元素，则返回 -1
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if i > j {
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return -1;
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}
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let m: i32 = (i + j) / 2;
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if nums[m as usize] < target {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
|
||
|
|
} else if nums[m as usize] > target {
|
||
|
|
// 递归子问题 f(i, m-1)
|
||
|
|
return dfs(nums, target, i, m - 1);
|
||
|
|
} else {
|
||
|
|
// 找到目标元素，返回其索引
|
||
|
|
return m;
|
||
|
|
}
|
||
|
|
}
|
||
|
|
|
||
|
|
/* 二分查找 */
|
||
|
|
fn binary_search(nums: &[i32], target: i32) -> i32 {
|
||
|
|
let n = nums.len() as i32;
|
||
|
|
// 求解问题 f(0, n-1)
|
||
|
|
dfs(nums, target, 0, n - 1)
|
||
|
|
}
|
||
|
|
```
|
||
|
|
|
||
|
|
=== "C"
|
||
|
|
|
||
|
|
```c title="binary_search_recur.c"
|
||
|
|
/* 二分查找：问题 f(i, j) */
|
||
|
|
int dfs(int nums[], int target, int i, int j) {
|
||
|
|
// 若区间为空，代表无目标元素，则返回 -1
|
||
|
|
if (i > j) {
|
||
|
|
return -1;
|
||
|
|
}
|
||
|
|
// 计算中点索引 m
|
||
|
|
int m = (i + j) / 2;
|
||
|
|
if (nums[m] < target) {
|
||
|
|
// 递归子问题 f(m+1, j)
|
||
|
|
return dfs(nums, target, m + 1, j);
|
||
|
|
} else if (nums[m] > target) {
|
||
|
|
// 递归子问题 f(i, m-1)
|
||
|
|
return dfs(nums, target, i, m - 1);
|
||
|
|
} else {
|
||
|
|
// 找到目标元素，返回其索引
|
||
|
|
return m;
|
||
|
|
}
|
||
|
|
}
|
||
|
|
|
||
|
|
/* 二分查找 */
|
||
|
|
int binarySearch(int nums[], int target, int numsSize) {
|
||
|
|
int n = numsSize;
|
||
|
|
// 求解问题 f(0, n-1)
|
||
|
|
return dfs(nums, target, 0, n - 1);
|
||
|
|
}
|
||
|
|
```
|
||
|
|
|
||
|
|
=== "Kotlin"
|
||
|
|
|
||
|
|
```kotlin title="binary_search_recur.kt"
|
||
|
|
/* 二分查找：问题 f(i, j) */
|
||
|
|
fun dfs(
|
||
|
|
nums: IntArray,
|
||
|
|
target: Int,
|
||
|
|
i: Int,
|
||
|
|
j: Int
|
||
|
|
): Int {
|
||
|
|
// 若区间为空，代表无目标元素，则返回 -1
|
||
|
|
if (i > j) {
|
||
|
|
return -1
|
||
|
|
}
|
||
|
|
// 计算中点索引 m
|
||
|
|
val m = (i + j) / 2
|
||
|
|
return if (nums[m] < target) {
|
||
|
|
// 递归子问题 f(m+1, j)
|
||
|
|
dfs(nums, target, m + 1, j)
|
||
|
|
} else if (nums[m] > target) {
|
||
|
|
// 递归子问题 f(i, m-1)
|
||
|
|
dfs(nums, target, i, m - 1)
|
||
|
|
} else {
|
||
|
|
// 找到目标元素，返回其索引
|
||
|
|
m
|
||
|
|
}
|
||
|
|
}
|
||
|
|
|
||
|
|
/* 二分查找 */
|
||
|
|
fun binarySearch(nums: IntArray, target: Int): Int {
|
||
|
|
val n = nums.size
|
||
|
|
// 求解问题 f(0, n-1)
|
||
|
|
return dfs(nums, target, 0, n - 1)
|
||
|
|
}
|
||
|
|
```
|
||
|
|
|
||
|
|
=== "Ruby"
|
||
|
|
|
||
|
|
```ruby title="binary_search_recur.rb"
|
||
|
|
[class]{}-[func]{dfs}
|
||
|
|
|
||
|
|
[class]{}-[func]{binary_search}
|
||
|
|
```
|
||
|
|
|
||
|
|
=== "Zig"
|
||
|
|
|
||
|
|
```zig title="binary_search_recur.zig"
|
||
|
|
[class]{}-[func]{dfs}
|
||
|
|
|
||
|
|
[class]{}-[func]{binarySearch}
|
||
|
|
```
|
||
|
|
|
||
|
|
??? pythontutor "Code Visualization"
|
||
|
|
|
||
|
|
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28nums%3A%20list%5Bint%5D,%20target%3A%20int,%20i%3A%20int,%20j%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%9A%E9%97%AE%E9%A2%98%20f%28i,%20j%29%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%EF%BC%8C%E4%BB%A3%E8%A1%A8%E6%97%A0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20-1%0A%20%20%20%20if%20i%20%3E%20j%3A%0A%20%20%20%20%20%20%20%20return%20-1%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%0A%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%92%E5%BD%92%E5%AD%90%E9%97%AE%E9%A2%98%20f%28m%2B1,%20j%29%0A%20%20%20%20%20%20%20%20return%20dfs%28nums,%20target,%20m%20%2B%201,%20j%29%0A%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%92%E5%BD%92%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i,%20m-1%29%0A%20%20%20%20%20%20%20%20return%20dfs%28nums,%20target,%20i,%20m%20-%201%29%0A%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%E5%85%B6%E7%B4%A2%E5%BC%95%0A%20%20%20%20%20%20%20%20return%20m%0A%0Adef%20binary_search%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%22%22%22%0A%20%20%20%20n%20%3D%20len%28nums%29%0A%20%20%20%20%23%20%E6%B1%82%E8%A7%A3%E9%97%AE%E9%A2%98%20f%280,%20n-1%29%0A%20%20%20%20return%20dfs%28nums,%20target,%200,%20n%20-%201%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%EF%BC%89%0A%20%20%20%20index%20%3D%20binary_search%28nums,%20target%29%0A%20%20%20%20print%28%22%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%206%20%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%22,%20index%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
||
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28nums%3A%20list%5Bint%5D,%20target%3A%20int,%20i%3A%20int,%20j%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%9A%E9%97%AE%E9%A2%98%20f%28i,%20j%29%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%EF%BC%8C%E4%BB%A3%E8%A1%A8%E6%97%A0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20-1%0A%20%20%20%20if%20i%20%3E%20j%3A%0A%20%20%20%20%20%20%20%20return%20-1%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%0A%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%92%E5%BD%92%E5%AD%90%E9%97%AE%E9%A2%98%20f%28m%2B1,%20j%29%0A%20%20%20%20%20%20%20%20return%20dfs%28nums,%20target,%20m%20%2B%201,%20j%29%0A%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%92%E5%BD%92%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i,%20m-1%29%0A%20%20%20%20%20%20%20%20return%20dfs%28nums,%20target,%20i,%20m%20-%201%29%0A%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%E5%85%B6%E7%B4%A2%E5%BC%95%0A%20%20%20%20%20%20%20%20return%20m%0A%0Adef%20binary_search%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%22%22%22%0A%20%20%20%20n%20%3D%20len%28nums%29%0A%20%20%20%20%23%20%E6%B1%82%E8%A7%A3%E9%97%AE%E9%A2%98%20f%280,%20n-1%29%0A%20%20%20%20return%20dfs%28nums,%20target,%200,%20n%20-%201%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%EF%BC%89%0A%20%20%20%20index%20%3D%20binary_search%28nums,%20target%29%0A%20%20%20%20print%28%22%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%206%20%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%22,%20index%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
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