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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.
Given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length under the problem constraint.
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
public class Solution { /** * @param nums: an array of integers * @param s: an integer * @return: an integer representing the minimum size of subarray */ public int minimumSize(int[] nums, int s) { // write your code here if(nums == null || nums.length == 0) return -1; int fast = 0; int slow = 0; int sum = 0; int result = Integer.MAX_VALUE; int size = nums.length; for(int i = 0; i < size; i++) sum += nums[i]; if(sum < s) return -1; sum = 0; while(fast < size){ while(fast < size && sum < s){ sum += nums[fast]; fast++; } //result = Math.min(result, fast - slow + 1); while(slow < size && sum >= s){ sum -= nums[slow]; slow++; } result = Math.min(result, fast - slow + 1); } return result; } }
lintcode-medium-Minimum Size Subarray Sum
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原文地址：http://www.cnblogs.com/goblinengineer/p/5341902.html
