LeetCode 209. Minimum Size Subarray Sum

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.

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.

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More practice:

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

分析：

运用前缀和，i到j之间的和为sum[j]-sum[i]+nums[i]，如果大于等于s，则与minn比较，求最小值

class Solution {public:    int minSubArrayLen(int s, vector<int>& nums) {                int minn=100000008;        int sum[nums.size()+1];        for(int i=0;i<nums.size();i++){            if(nums[i]>=s) return 1;            if(i==0) sum[i]=nums[i];            else sum[i]=sum[i-1]+nums[i];        }        for(int i=0;i<nums.size();i++){            for(int j=i+1;j<nums.size();j++){                if(sum[j]-sum[i]+nums[i]>=s){                    minn=min(j-i+1,minn);                    if(minn==2) return 2;                    break;                }            }        }        if(minn==100000008) return 0;        else return minn;    }};