代码: publicintminSubArrayLen(ints,int[] nums) {if(nums==null|| nums.length<=0)return0;intlen=nums.length;intans=Integer.MAX_VALUE;intslow=0,fast=0;intsum=0;while(fast<len){ sum+=nums[fast++];while(sum>=s){//前面sum与nums[fast++]相加,fast也自加了1,所以比较min与fast-slow即可...
classSolution{public:intminSubArrayLen(ints,vector<int>& nums){intlen = nums.size();if(len ==0)return0;// 数组为空intminlen = INT_MAX;intsum =0;intleft =0;// 滑动窗口的左右边沿intright =-1;while(right < len) {while(sum < s && right < len) sum += nums[++right];// 向右...
Leetcode力扣 209 | 长度最小的子数组 Minimum Size Subarray Sum 2020年11月05日 07:455743浏览·30点赞·11评论 爱学习的饲养员 粉丝:7.0万文章:46 关注 视频讲解 622:17 Leetcode力扣 1-300题视频讲解合集|手画图解版+代码【持续更新ing】 84.4万798 ...
[Leetcode] 209. Minimum Size Subarray Sum link: https://leetcode.com/problems/minimum-size-subarray-sum/ Solution: 双指针 需要注意左边指针右移次数的判定 TC: O(n) SC: O(1)...LeetCode 209. Minimum Size Subarray Sum LeetCode 209. Minimum Size Subarray Sum Description Given an array ...
LeetCode: 209. 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&r...Leetcode 209. Minimum Size Subarray Sum Given an array of n positive integers...
定义两个指针left和right,分别记录子数组的左右的边界位置,然后我们让right向右移,直到子数组和大于等于给定值或者right达到数组末尾,此时我们更新最短距离,并且将left像右移一位,然后再sum中减去移去的值,然后重复上面的步骤,直到right到达末尾,且left到达临界位置,即要么到达边界,要么再往右移动,和就会小于给定值。
func minSubArrayLen(s int, nums []int) int { if len(nums)==0 { return 0 } l := 0 r := 0 curWs := nums[0] curWz := 1 endIndex := len(nums) - 1 arrLen := len(nums) curMinSize := arrLen //for r < endIndex ||( l <= r && l<endIndex) { for l!=endIndex ...
每天一算:Minimum Size Subarray Sum leetcode上第209号问题:Minimum Size Subarray Sum 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。 示例: 输入: s = 7, nums = [2,3,1,2,4,3]...
LeetCode-209. Minimum Size Subarray Sum Given an array ofnpositive integers and a positive integers, find the minimal length of acontiguoussubarray 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...
}if(left == n +1)break; res= min(res, left -i); }returnres == INT_MAX ?0: res; } }; 本文转自博客园Grandyang的博客,原文链接:最短子数组之和[LeetCode] Minimum Size Subarray Sum,如需转载请自行联系原博主。