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1f784dadb0
divide and conquer.
46 lines
No EOL
1.1 KiB
C++
46 lines
No EOL
1.1 KiB
C++
/**
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* File: binary_search_recur.cpp
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* Created Time: 2023-07-17
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* Author: krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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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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/* Driver Code */
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int main() {
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int target = 6;
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vector<int> nums = {1, 3, 6, 8, 12, 15, 23, 26, 31, 35};
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// 二分查找(双闭区间)
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int index = binarySearch(nums, target);
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cout << "目标元素 6 的索引 = " << index << endl;
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return 0;
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} |