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; } };