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58 lines
1.5 KiB
Markdown
58 lines
1.5 KiB
Markdown
# [560. Subarray Sum Equals K](https://leetcode.com/problems/subarray-sum-equals-k/)
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## 题目
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Given an array of integers `nums` and an integer `k`, return *the total number of continuous subarrays whose sum equals to `k`*.
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**Example 1:**
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```
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Input: nums = [1,1,1], k = 2
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Output: 2
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```
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**Example 2:**
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```
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Input: nums = [1,2,3], k = 3
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Output: 2
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```
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**Constraints:**
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- `1 <= nums.length <= 2 * 104`
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- `-1000 <= nums[i] <= 1000`
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- `-10^7 <= k <= 10^7`
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## 题目大意
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给你一个整数数组 `nums` 和一个整数 `k` ，请你统计并返回该数组中和为 `k` ****的连续子数组的个数。
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## 解题思路
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- 此题不能使用滑动窗口来解。因为 `nums[i]` 可能为负数。
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- 前缀和的思路可以解答此题，但是时间复杂度有点高了，`O(n^2)`。考虑优化时间复杂度。
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- 题目要求找到连续区间和为 `k` 的子区间总数，即区间 `[i,j]` 内的和为 K ⇒ `prefixSum[j] - prefixSum[i-1] == k`。所以 `prefixSum[j] == k - prefixSum[i-1]` 。这样转换以后，题目就转换成类似 A + B = K 的问题了。LeetCode 第一题的优化思路拿来用。用 map 存储累加过的结果。如此优化以后，时间复杂度 `O(n)`。
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## 代码
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```go
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package leetcode
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func subarraySum(nums []int, k int) int {
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count, pre := 0, 0
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m := map[int]int{}
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m[0] = 1
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for i := 0; i < len(nums); i++ {
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pre += nums[i]
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if _, ok := m[pre-k]; ok {
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count += m[pre-k]
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}
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m[pre] += 1
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}
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return count
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}
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``` |