Given an array of integers nums and an integer k, return the total number of continuous subarrays whose sum equals to k.

 

Example 1:

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

Example 2:

Input: nums = [1,2,3], k = 3
Output: 2

哦，看错题了。是数组的个数，不是数组的长度。
把超出的元素去掉，就符合了。所以判断的是超出元素出现的次数

 

Your input
[1,1,1]
2
stdout
sum = 1
sum = 2
超出的元素sum - k = 0
超出的元素出现的次数preSum.get(sum - k) = 1
 
sum = 3
超出的元素sum - k = 1
超出的元素出现的次数preSum.get(sum - k) = 1

 

public class Solution {
    public int subarraySum(int[] nums, int k) {
        int sum = 0, result = 0;
        Map<Integer, Integer> preSum = new HashMap<>();
        preSum.put(0, 1);
        
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            System.out.println("sum = " + sum);
            
            if (preSum.containsKey(sum - k)) {
                result += preSum.get(sum - k);
                System.out.println("超出的元素sum - k = " + (sum - k));
                System.out.println("超出的元素出现的次数preSum.get(sum - k) = " + preSum.get(sum - k));
                System.out.println(" ");
            }
            preSum.put(sum, preSum.getOrDefault(sum, 0) + 1);
        }
        
        return result;
    }
}

View Code