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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 -1 instead.
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.
分析：
这里类似于在array上面放一个window， window的左边从第一个数开始，增加window的size, 保证window里的数之和大于S，然后每当window 右边往前走一步（window变大），就check是否可以把window左边cover的值remove掉（window 变小），并同时update minLength.
1 public class Solution { 2 /** 3 * @param nums: an array of integers 4 * @param s: an integer 5 * @return: an integer representing the minimum size of subarray 6 * cnblogs.com/beiyeqingteng/ * 7 */ 8 public int minimumSize(int[] nums, int s) { 9 if (nums == null || nums.length == 0) return -1; 10 int start = 0; 11 int total = 0; 12 int minLength = Integer.MAX_VALUE; 13 for (int end = 0; end < nums.length; end++) { 14 total += nums[end]; 15 if (total == s && minLength == Integer.MAX_VALUE) { 16 minLength = end - start + 1; 17 } 18 while (start <= end && total - nums[start] >= s ) { 19 total -= nums[start]; 20 start++; 21 if (minLength > end - start + 1) { 22 minLength = end - start + 1; 23 } 24 } 25 } 26 27 if (total < s) return -1; 28 return minLength; 29 } 30 }
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5631737.html
