【LeetCode】15.Array and String —Minimum Size Subarray Sum 满足条件和的最小子串

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). 

 

学习计算机网络的时候，get到了一种滑动窗口的思想，对于此类寻找连续字串的题目有很好的参考意义。首先从数组的最左边（最开始）一词累加，知道累加和大于等于输入值，然后从左边开始减少（考虑到数组是无序的），小于输入值则向右滑动。想了很久，才想到。。。

class Solution {
public:
int minSubArrayLen(int s, vector<int>&nums)
{
    //滑动窗口的思想
    int size = nums.size();//获取数组长度
    int result=INT_MAX;//定义返回结果
    int Sum = 0; //计算总和，用于比较
    int begin = 0;//滑动窗口开始
    for (int i = 0; i < size; i++)
    {
        Sum += nums[i]; //向数组右端开始累加
        while (Sum >= s) {//满足要求时向右滑动，改变窗口开始begin的值
            result = min(result, i - begin + 1);
            Sum -= nums[begin];
            begin++;
        }
    }
    return result == INT_MAX ? 0 : result;
}
};