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:
题目含义：求和为k的连续子数组的个数
1 public int subarraySum(int[] nums, int k) { 2 int[] preSum = new int[nums.length]; //dp数组，位置i上存储从0到i个元素总和 3 preSum[0] = nums[0]; 4 for (int i = 1; i < nums.length; i++) preSum[i] = preSum[i - 1] + nums[i]; 5 int result = 0; 6 for (int i = 0; i < preSum.length; i++) { 7 if (preSum[i] == k) result++; 8 for (int j = i + 1; j < preSum.length; j++) { 9 if (preSum[j] - preSum[i] == k) result++; //线段2-线段1==长度k 10 } 11 } 12 return result; 13 }
原文：http://www.cnblogs.com/wzj4858/p/7678443.html