209. Minimum Size Subarray Sum 解题记录

题目描述：

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.

解题思路：

第一次做的时候我的思路就是寻找每一个子数组，找到最少元素的那一个。

第二次稍微参考了下别人的思路，重新整理了下代码，这次的思路是相当于用一个动态的子数组在数组上搜索，分别用left和right作为子数组的边界，当子数组中元素和大于等于s时：比较长度，做左边界右移；小于时：右边界右移。

第一次代码：

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int n = nums.size(), minNum = INT_MAX;
        for(int i = 0; i < n; i++){
            int tSum = 0, j = i;
            for(; j < n; j++){
                tSum += nums[j];
                if(tSum >= s)
                    break;
            }
            if(tSum >= s)
                minNum = min(j-i+1, minNum);
        }
        if(minNum == INT_MAX)
            return 0;
        return minNum;
    }
};

 第二次代码：

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int n = nums.size(), minNum = INT_MAX, left = 0, right = 0;
        int sum = 0;
        while(right <= n && left <= right){
        //当右边界越界的时候，最后一个元素作为右边界的情况还未遍历，所以时<=
            if(sum < s){
            //小于
                if(right == n)
                //最后一个元素是右边界的终止条件
                    break;
                sum += nums[right++];
            }
            else{
            //大于等于
                minNum = min(right-left, minNum);
                sum -= nums[left++];
            }
        }
        if(minNum == INT_MAX)
        //不存在的情况
            return 0;
        return minNum;
    }
};