数组连续子数组求和的问题

这是一道非常经典的题，这道题是求出所有连续子数组中求和大于等于某一个给定值的长度最短的那个连续子数组的长度。其实就是Two-pointer问题，我们通过while（注意pointer条件），一般两层while就可以，一个rightindex自增，另一个leftindex自增，即可实现。

209. Minimum Size Subarray Sum

Medium

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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). 

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int sum = 0;
        int LeftIndex = 0;
        int RightIndex = 0;
        int Result = 1e8;
        while(RightIndex < nums.size()){
            while(sum < s && RightIndex < nums.size()){
                sum += nums[RightIndex++];
            }
            while(sum >= s && LeftIndex < nums.size()){
                Result = min(Result,RightIndex - LeftIndex);
                sum -= nums[LeftIndex++];
            }
        }
        if(RightIndex == nums.size() && LeftIndex == 0){
            Result = 0;
        }
        return Result;
    }
};