【LeetCode-209】Minimum Size Subarray Sum

咋一看挺麻烦，其实就是一层膜，捅破它你就恍然大明白了，开个玩笑，上代码吧，时间复杂度O(n)

public class MinimumSizeSubarraySum {public int minSubArrayLen(int s, int[] nums) {if (nums.length == 0) {return 0;}int result = Integer.MAX_VALUE;int begin = 0,sum = 0;for (int i = 0; i < nums.length; i ++) {sum += nums[i];if (sum >= s) {// 当前的数的个数与最后的结果进行比较int currentCount = i - begin + 1;if (currentCount < result) {result = currentCount;}//当前的和减去nums[begin]和nums[i],因为下一次循环还要加nums[i]sum -= nums[i];sum -= nums[begin];i --;begin ++;} }return result == Integer.MAX_VALUE ? 0 : result;}}