【leetcode】Minimum Size Subarray Sum（middle）

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.

思路：用start, end两个游标来记录范围，sum < s end就向后走， s >= sum start就向后走。

我写的代码没有大神的逻辑清晰，先上大神的。

int minSubArrayLen(int s, vector<int>& nums) {
    int firstPos = 0, sum = 0, minLength = INT_MAX;
    for(int i = 0; i<nums.size(); i++) {   //i即end游标 对所有end游标循环
        sum += nums[i];
        while(sum >= s) {   //对每个end游标的start游标循环 firstPos即为start游标 只有s >= sum 时才把start向后移
            minLength = min(minLength, i - firstPos + 1);
            sum -= nums[firstPos++];
        }
    }

    return minLength == INT_MAX? 0 : minLength;  //没找到s >= sum 时返回0
  }

我的代码乱一点，但是也AC了。

int minSubArrayLen(int s, vector<int>& nums) {
        int start = 0, end = 0;
        int sum = 0;
        int minLength = nums.size() + 1;
        while(end <= nums.size()) //有等于是因为结尾到最后面时 起始点还可能移动
        {
            if(sum < s)
            {
                if(end == nums.size()) break;
                sum += nums[end++];
            }
            else
            {
                minLength = (minLength < (end - start)) ? minLength : (end - start);
                sum -= nums[start++];
            }
        }
        minLength = (minLength == nums.size() + 1) ? 0 : minLength; //没找到符合条件的子序列 返回0
        return minLength;
    }