209. Minimum Size Subarray Sum**

209. Minimum Size Subarray Sum**

https://leetcode.com/problems/minimum-size-subarray-sum/

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.

Example:

Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

Follow up:
If you have figured out the $O(n)$ solution, try coding another solution of which the time complexity is $O(n\log n)$.

C++ 实现 1

双指针. 查找 nums[i ... j) 之和 $\text{sum} \ge s$. 我下面的代码有点魔幻~.

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

C++ 实现 2

来自 LeetCode Submission. 这个代码可能更符合常规.

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