209. Minimum Size Subarray Sum

209. Minimum Size Subarray Sum

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Total Accepted: 26840  Total Submissions: 105654  Difficulty: Medium

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.

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Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

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分析：

这个题目一来就应该想到应该一点一点累加恰好找到这个s的累加和，统计出长度
所以接着缩小数据，再次找到恰好大于s的位置，更新最小长度，恰好是为了尽可能小
具体：用一个迭代器慢ite1，一个迭代器快ite2
如果当前序列的和小于s则ite2继续右移动（增加和），否则ite1右移动（相当于减小和）

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        if(nums.empty())
            return 0;
        vector<int>::iterator ite1=nums.begin();
        vector<int>::iterator ite2=ite1;
        int minLen=INT_MAX;
        int curSum=*ite1,curLen=1;
        while(true)
        {
            if(curSum >= s )
            {   
                if(curLen <minLen)
                    minLen=curLen;
                curSum-=*ite1;
                ite1++;
                curLen--;
                continue;
            }
            ite2++;
            if(ite2==nums.end())
                break;
            curLen++;
            curSum+=*ite2;
        }
        
        if(minLen==INT_MAX)
            return 0;
        return minLen;    
    }
};

别人的算法

（本质一样），O(N)的速度：

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int n = nums.size(), start = 0, sum = 0, minlen = INT_MAX;
        for (int i = 0; i < n; i++) { 
            sum += nums[i]; 
            while (sum >= s) {
                minlen = min(minlen, i - start + 1);
                sum -= nums[start++];
            }
        }
        return minlen == INT_MAX ? 0 : minlen;
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50453943

原作者博客：http://blog.csdn.net/ebowtang