leetcode 209. Minimum Size Subarray Sum

最新推荐文章于 2024-06-16 11:54:10 发布

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最新推荐文章于 2024-06-16 11:54:10 发布

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本文介绍了一种寻找具有最小长度且其元素总和大于或等于给定值s的连续子数组的方法。通过两种不同的算法实现：一种是基础的O(n^2)方法，另一种是使用两个指针进行移动的高效算法。

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209. Minimum Size Subarray Sum

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.

法一：最基础的方法 o(n^2)

法二：两个指针，移动

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) 
    {
        //way-1 有一部分重复运算
        /*
        int min = nums.size() + 1;
        int sum = 0;
        for (int i = 0; i < nums.size(); i++)
        {
            for (int j = i; j < nums.size(); j++)
            {
                sum += nums[j];
                if(sum >= s && j - i + 1 < min)
                    min = j - i + 1;
            }
            sum = 0;
        }
        if (min == nums.size() + 1) 
            return 0; 
        return min; 
        */
        
        //way-2
        int first = 0, second = -1;
        int sum = 0;
        int length = nums.size();
        int ret = length + 1;
        while (first < length && second < length)
        {
            //second后移，直到sum超过s 
            while(sum < s && second < length)
            {
                second++;
                sum += nums[second];
            }       
           
            if(second == nums.size())
                break;
              
            while(sum >= s)      
            {
                ret = min(ret, (second-first+1));
                sum -= nums[first];
                first++;
            }      
        } 
        if (ret == nums.size() + 1)
            return 0;
        return ret;
    }
};