1classSolution {2public:3intminSubArrayLen(ints, vector<int>&nums) {4intsz =nums.size();5vector<int>ret;6if(sz ==0)return0;7inti =0;8intj =0;9inttmpSum =0;10while(j <sz){11for( ; i < sz; ++i){12tmpSum +=nums[i];13if(tmpSum >=s)14break;15}16if(tmpSum < s)br...
classSolution{publicintminSubArrayLen(inttarget,int[] nums){intlength=nums.length;if(length <=0) {return0; }intsubArrayLength=Integer.MAX_VALUE;intstart=0;intend=0;intsum=0;while(end < nums.length) { sum += nums[end];while(sum >= target) { subArrayLength = Math.min(subArrayLength, ...
Explanation: the subarray [4,3] has the minimal length under the problem constraint. 1. 2. 3. 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). 分析 题目的意思是:求连续子数组只和等于s的最小长度。 定义两...
题目地址: https://leetcode.com/problems/minimum-size-subarray-sum/description/ 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...
publicclassSolution{/** * @param nums: an array of integers * @param s: an integer * @return: an integer representing the minimum size of subarray */publicintminimumSize(int[]nums,int s){int[]sums=newint[nums.length+1];for(int i=1;i<sums.length;i++)sums[i]=sums[i-1]+nums[...
class Solution { public: int minSubArrayLen(int s, vector<int>& nums) { int res=INT_MAX; int sum=0; for(int i=0,j=0;i<nums.size();){ sum+=nums[i]; i++; while(sum>=s){ res=min(res,i-j); sum-=nums[j]; j++; ...
1// 209. Minimum Size Subarray Sum 2// https://leetcode.com/problems/minimum-size-subarray-sum/description/ 3// 4// 滑动窗口的思路 5// 时间复杂度: O(n) 6// 空间复杂度: O(1) 7classSolution{ 8public: 9intminSubArrayLen(ints,vector<int>& nums){ ...
}if(left == n +1)break; res= min(res, left -i); }returnres == INT_MAX ?0: res; } }; 最短子数组之和[LeetCode] Minimum Size Subarray Sum,如需转载请自行联系原博主。
class Solution { public int minSubArrayLen(int s, int[] nums) { int left = 0, right = 0, sum = 0, res = Integer.MAX_VALUE; while (left <=right && right <= nums.length) { if (sum < s) { if (right <nums.length) { ...
LeetCode 209. Minimum Size Subarray Sum 简介:给定一个含有 n 个正整数的数组和一个正整数 s ,找出该数组中满足其和 ≥ s 的长度最小的连续子数组。如果不存在符合条件的连续子数组,返回 0。 Description Given an array of n positive integers and a positive integer s, find the minimal length of a...