[leetcode]560. Subarray Sum Equals K 和为K的子数组

和为K的子数组算法解析

最新推荐文章于 2025-09-07 20:32:33 发布

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最新推荐文章于 2025-09-07 20:32:33 发布

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原文链接：http://www.cnblogs.com/liuliu5151/p/9810120.html

本文详细解析了一种高效查找数组中和为特定数值K的连续子数组数量的算法。通过使用哈希表记录前缀和的频率，算法能在O(n)的时间复杂度内完成任务，适用于长度高达20,000的数组。文章还提供了完整的Java代码实现，帮助读者深入理解算法原理。

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1:

Input:nums = [1,1,1], k = 2
Output: 2

Note:

The length of the array is in range [1, 20,000].
The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

题目

和为K的子数组

思路：

1. The key to solve this problem is to find subarray sum[i, j] == k

2. if we know sum[0, i-1], sum[0,j] we can check sum[0, i-1] - k == sum[0,j] ?

3. To achieve this, we traversal whole array, save sum[0,j] as preSum in map, checking curSum[0,i-1] - k is contained in map

nums = [5, 　　2, 　　3, 　　4, 　　5] k= 7

sum=5

=10

=14

map

sum frequency check (sum - k) in map?

init 0 1

5 1 5-7=-2 No

7 1 7-7=0 Yes update res

10 1 10-7=3 No

14 1 14-7=7 Yes update res

代码：

 1 public class Solution {
 2     public int subarraySum(int[] nums, int k) {
 3         int sum = 0, result = 0;
 4         Map<Integer, Integer> map = new HashMap<>();
 5         map.put(0, 1);
 6         
 7         for (int i = 0; i < nums.length; i++) {
 8             sum += nums[i];
 9             if (map.containsKey(sum - k)) {
10                 result += map.get(sum - k);
11             }
12             map.put(sum, map.getOrDefault(sum, 0) + 1);
13         }
14         
15         return result;
16     }
17 }