2022-11-22 17:47:26 +08:00
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# 权衡时间与空间
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理想情况下,我们希望算法的时间复杂度和空间复杂度都能够达到最优,而实际上,同时优化时间复杂度和空间复杂度是非常困难的。
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**降低时间复杂度,往往是以提升空间复杂度为代价的,反之亦然。** 我们把牺牲内存空间来提升算法运行速度的思路称为「以空间换时间」;反之,称之为「以时间换空间」。选择哪种思路取决于我们更看重哪个方面。大多数情况下,内存空间不会成为算法瓶颈,因此以空间换时间更加常用。
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## 示例题目 *
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以 LeetCode 全站第一题 [两数之和](https://leetcode.cn/problems/two-sum/) 为例,「暴力枚举」和「辅助哈希表」分别为 **空间最优** 和 **时间最优** 的两种解法。本着时间比空间更宝贵的原则,后者是本题的最佳解法。
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### 方法一:暴力枚举
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时间复杂度 $O(N^2)$ ,空间复杂度 $O(1)$ ,属于「时间换空间」。
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虽然仅使用常数大小的额外空间,但运行速度过慢。
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=== "Java"
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```java title="" title="leetcode_two_sum.java"
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public int[] twoSum(int[] nums, int target) {
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int size = nums.length;
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// 外层 * 内层循环,时间复杂度为 O(n)
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for (int i = 0; i < size - 1; i++) {
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for (int j = i + 1; j < size; j++) {
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if (nums[i] + nums[j] == target)
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return new int[] { i, j };
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}
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}
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return new int[0];
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}
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```
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=== "C++"
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```cpp title="leetcode_two_sum.cpp"
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2022-11-24 22:54:10 +08:00
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2022-11-22 17:47:26 +08:00
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```
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=== "Python"
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```python title="leetcode_two_sum.py"
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```
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2022-11-24 22:54:10 +08:00
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=== "Go"
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```go title="leetcode_two_sum.go"
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func twoSum(nums []int, target int) []int {
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size := len(nums)
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for i := 0; i < size-1; i++ {
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for j := i + 1; i < size; j++ {
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if nums[i]+nums[j] == target {
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return []int{i, j}
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}
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}
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}
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return nil
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}
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```
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2022-11-22 17:47:26 +08:00
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### 方法二:辅助哈希表
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时间复杂度 $O(N)$ ,空间复杂度 $O(N)$ ,属于「空间换时间」。
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借助辅助哈希表 dic ,通过保存数组元素与索引的映射来提升算法运行速度。
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=== "Java"
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```java title="" title="leetcode_two_sum.java"
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public int[] twoSum(int[] nums, int target) {
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int size = nums.length;
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// 辅助哈希表,空间复杂度 O(n)
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Map<Integer, Integer> dic = new HashMap<>();
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// 单层循环,时间复杂度 O(n)
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for (int i = 0; i < size; i++) {
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if (dic.containsKey(target - nums[i])) {
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return new int[] { dic.get(target - nums[i]), i };
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}
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dic.put(nums[i], i);
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}
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return new int[0];
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}
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```
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=== "C++"
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```cpp title="leetcode_two_sum.cpp"
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2022-11-24 22:54:10 +08:00
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2022-11-22 17:47:26 +08:00
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```
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=== "Python"
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```python title="leetcode_two_sum.py"
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```
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2022-11-24 22:54:10 +08:00
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=== "Go"
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```go title="leetcode_two_sum.go"
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func twoSumHashTable(nums []int, target int) []int {
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hashTable := map[int]int{}
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for idx, val := range nums {
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if preIdx, ok := hashTable[target-val]; ok {
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return []int{preIdx, idx}
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}
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hashTable[val] = idx
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}
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return nil
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}
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```
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