Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn‘t one, return 0 instead.
For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.
public class Solution { public int minSubArrayLen(int s, int[] nums) { //借助一个指针start，判断从start到i的sum是否大于s，如果大于，start++，并且不断更新minLen //注意，一开始需要将minLen初始化为Integer.MAX_VALUE int minLen=Integer.MAX_VALUE; int start=0; int sum=0; for(int i=0;i<nums.length;i++){ sum+=nums[i]; while(sum>=s){ minLen=Math.min(minLen,i-start+1); sum-=nums[start]; start++; } } return minLen==Integer.MAX_VALUE?0:minLen; } }
[leedcode 209] Minimum Size Subarray Sum
原文：http://www.cnblogs.com/qiaomu/p/4703356.html