LeetCode 209. Minimum Size Subarray Sum

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本文介绍了一种在给定正整数数组中寻找最小子数组的方法，该子数组的和大于或等于特定值s。通过O(n)的时间复杂度算法实现，同时探讨了O(nlogn)的解决方案。

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Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.

Example:

Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

Follow up:

If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

代码：

class Solution {
    public int minSubArrayLen(int s, int[] nums) {
       int l=0,r=-1; //sum[0,-1]
       int sum=0;
       int length=nums.length+1;
       while(l<nums.length-1) {
           if(r+1<nums.length&&sum<s){
               sum+=nums[++r];
           }else{
               sum-=nums[l++];
           }
           if(sum>=s)
               length=Math.min(length,r-l+1);
       }
       
        if(length==nums.length+1)
            return 0;
        return length;    
    }
}