209. Minimum Size Subarray Sum

问题描述:

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). 

解题思路:

可以用滑动窗口来做。

注意审题!!注意审题!!注意审题!!!!

sum ≥ s

O(nlogn) 会涉及二分法,见Grandyang

代码:

O(n):

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        if(nums.empty())
            return 0;
        int left = 0;
        int ret = nums.size()+1;
        int sum = 0;
        for(int i = 0; i < nums.size(); i++){
            sum += nums[i];
            while(sum >= s){
                ret = min(ret, i - left + 1);
                sum -= nums[left++];
            }
        }
        return ret == nums.size() + 1 ? 0 : ret;
    }
};
原文地址:https://www.cnblogs.com/yaoyudadudu/p/9197851.html