leetcode 209. Minimum Size Subarray Sum 最短子数组的和 + 十分典型的移动窗口做法

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.

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More practice:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

寻找大于target的最短的子数组。

最自觉的方法肯定超时，这里有一个类似滑动窗口的方法，就是双指针的方法，起初我也没想出来，所以先记下吧！

要和这一道题leetcode 53. Maximum Subarray DP+最大子串和 一起学习。

这是十分明显的移动窗口的做法，十分值得学习，

代码如下：

/*
 * 双指针来解决这个问题
 * http://www.cnblogs.com/grandyang/p/4501934.html
 * 
 * 很好地解法，需要继续学习
 * 
 * */

public class Solution {
    public int minSubArrayLen(int s, int[] nums) 
    {
        if(nums==null || nums.length<=0)
            return 0;

        int left=0,right=0,res=Integer.MAX_VALUE;
        int sum=0;
        while(rightwhile(rightsumsum+=nums[right++];
            while(sum>=s)
            {
                res = Math.min(res, right-left);
                sum-=nums[left++];
            }
        }
        return res==Integer.MAX_VALUE? 0 : res;
    }
}

下面是C++的做法，这道题用到了移动窗口的思想，一次遍历即可解决，很棒的做法，很值得学习

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

class Solution
{
public:
    int minSubArrayLen(int s, vector<int>& n) 
    {
        if (n.size() <= 0)
            return 0;

        int left = 0, right = 0, sum = 0;
        int res = numeric_limits<int>::max();
        while (right < n.size())
        {
            while (right < n.size() && sum < s)
                sum += n[right++];
            while (left<=right && sum>=s)
            {
                res = min(res, right - left);
                sum -= n[left++];
            }
        }
        return res == numeric_limits<int>::max() ? 0 : res;
    }
};