* @return: an integer representing the minimum size of subarray */ publicintminimumSize(int[] nums,ints) { intans = Integer.MAX_VALUE; inti, j =0; intsum =0; //for 循环主指针 for(i =0; i < nums.length; i++) { sum += nums[i]; // while 循环辅指针 while(j < nums.lengt...
intLeetCode::minSubArrayLen3(ints, vector<int>&nums){intsum =0, ret = nums.size() +1;//ret记录最小值vector<int>& sums(nums);//前i项的和for(inti =0; i < nums.size(); i++) sums[i]= nums[i] + (i ==0?0: sums[i -1]);for(inti =0; i < nums.size(); i++) {...
int[][] subArrayItemSum = getSubArrayItemSum(nums); if(subArrayItemSum == null){ return 0; } //找到符合条件的最短子数组 for(int i=0; i<subArrayItemSum.length;i++){ if(subArrayItemSum[i][0] >=s){ if(subArrayItemSum[i][1] < length){ length = subArrayItemSum[i][1]; } }...
publicintminSubArrayLen(ints,int[]nums){intn=nums.length;if(n==0){return0;}intleft=0;intright=0;intsum=0;intmin=Integer.MAX_VALUE;while(right<n){sum+=nums[right];right++;while(sum>=s){min=Math.min(min,right-left);sum-=nums[left];left++;}}returnmin==Integer.MAX_VALUE?0:min...
int minSubArrayLen(int s, vector<int>& nums) { if(nums.empty()){ return 0; } int i=0; int j=i+1; int sum=nums[i]; int m=nums.size(); int len=m+1; while(j<m){ while(sum<s&&j<m){ sum+=nums[j]; j++; }
(2)更新最短距离,将left像右移一位,sum减去移去的值; (3)重复(1)(2)步骤,直到right到达末尾,且left到达临界位置 动画演示 代码 1// 209. Minimum Size Subarray Sum 2// https://leetcode.com/problems/minimum-size-subarray-sum/description/ ...
这道题给定了我们一个数字,让我们求子数组之和大于等于给定值的最小长度,跟之前那道Maximum Subarray有些类似,并且题目中要求我们实现 O(n) 和 O(nlgn) 两种解法,那么我们先来看 O(n) 的解法,我们需要定义两个指针 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...
LeetCode 209: Minimum Size Subarray Sum(长度最小的子数组) Q: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. ...
The computational overhead mainly results from the L脳L covariance matrix inversion needed for computation of the adaptive weights, the complexity of which is cubic with the subarray size, O(L3). In medical ultrasound imaging with focusing on the imaging point, we have a limited number of ...
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...