标签：log tput who span 存储 amp bsp ota 长度
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 }
标签：log tput who span 存储 amp bsp ota 长度
原文地址：http://www.cnblogs.com/wzj4858/p/7678443.html
