[Leetcode 209, Medium] Minimum Size Subarray Sum-优快云博客

本文探讨了一个算法问题：给定一个数组和一个正整数s，找到数组中满足和大于等于s的最短子数组长度。提供了解决方案并分析了其动态规划性质，还给出了C++、Java和Python的实现代码。

Problem:

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.

click to show more practice.

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).

Analysis:

Dynamic programming.

Solutions:

C++:

int minSubArrayLen(int s, vector<int>& nums) {
    if(nums.empty())
        return 0;

    int size = nums.size();
    int sum = nums[0];
    if(sum >= s)
        return 1;

    int start = 0;
    int end = 1;
    int min_length_of_subarray = -1;
    while(start <= end) {
        if(sum >= s) {
            if(start == end - 1)
                return 1;
            else if(min_length_of_subarray == -1 || min_length_of_subarray > end - start)
                min_length_of_subarray = end - start;

            sum -= nums[start++];
        } else {
            if(end == size)
                break;
            sum += nums[end++];
        }
    }

    return min_length_of_subarray == -1 ? 0 : min_length_of_subarray;
}

Java :

Python: