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

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:

1. The length of the array is in range [1, 20,000].
2. 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

=7

=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：

 public class Solution {
     public int subarraySum(int[] nums, int k) {
         int sum = 0, result = 0;
         Map<Integer, Integer> map = new HashMap<>();
         map.put(0, 1);

         for (int i = 0; i < nums.length; i++) {
             sum += nums[i];
             if (map.containsKey(sum - k)) {
                 result += map.get(sum - k);
             }
             map.put(sum, map.getOrDefault(sum, 0) + 1);
         }

         return result;
     }
 }