[Leetcode]209. Minimum Size Subarray Sum

Problem

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.

Solution

维护一个窗口，该窗口尽量使得内部的元素之和大于等于 S， 

这样如果小于 s , 就移动又窗口，

否则就移动左窗口

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        if(nums.empty() || s < 0) return 0;
        
        const int N = nums.size();
        int rst = INT_MAX, curSum = 0;
        for( int left = 0, right = 0; right <= N; ){
            if( curSum < s){
                if(right == N) break;
                curSum += nums[right++];
            }
            else {
                rst = min( rst, right - left);
                curSum -= nums[left++];
            }
        }
        return rst==INT_MAX ? 0 : rst;
    }
};

Bug ： right <= nums.size()  是停止条件

Solution 2

因为全部都是正数，可以累加起来，然后用二分查找。