1publicclassSolution {2publicintminSubArrayLen(ints,int[] nums) {3if(nums ==null|| nums.length == 0)4return0;5intlength = nums.length, left = 0, right = 0, result = length, sum = nums[0];6while(right <length) {7if(left ==right) {8if(nums[left] >=s) {9return1;10}els...
1classSolution {2public:3intminSubArrayLen(ints, vector<int>&nums) {4intn = nums.size(), minNum = INT_MAX, left =0, right =0;5intsum =0;6while(right <= n && left <=right){7//当右边界越界的时候,最后一个元素作为右边界的情况还未遍历,所以时<=8if(sum ...
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]; } }...
left和right最初都指向0; right右移直到sum>=s;再右移left直到sum<s,之间更新res;重复上述两个过程。 时间复杂度O(n)。 class Solution { public: int minSubArrayLen(int s, vector<int>& nums) { int left=0,right=0; int len=nums.size(); int res=len+1; int sum=0; while(right<len){ w...
www.lintcode.com/en/problem/minimum-size-subarray-sum/ 【题目解析】 sum为前i个数的和,长度比nums多一个,sum[0] = 0。这样从0开始一直到len,遍历sum计算sum[j]与sum[i]的差大于等于s的时候的j-i长度,把它与minlen比较,如果比minlen小就更新minlen。
classSolution2{publicintminSubArrayLen(ints,int[]nums){if(nums==null||nums.length==0)return0;// integral: accumulate sumintsums[]=newint[nums.length+1];for(inti=1;i<sums.length;i++){sums[i]=sums[i-1]+nums[i-1];}// binar search, for every sums[i],// find the sum[end] ...
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...
Sub-array size: 4 Sub-array from 0 to 3 and sum is: 10 Sample Solution: Scala Code: objectScala_Array{deffind_min_subarray_sum(nums:Array[Int],k:Int):Array[Int]={varsub_arr_sum=0;varmin_sub_arr=Integer.MAX_VALUE;varlast=0;varresult=newArray[Int](3)for(i<-0tonums.length-1)...
Leetcode 209. Minimum Size Subarray Sum 0 / 0 / 创建于 5年前 / 复盘 没有明确 循环终止条件和循环终止状态定义 需要走几个极端的测试用例 复盘: 对比bobo sir 的定义: 初始l=0 , r=-1 , curMin=len +1 初始化的情况要特殊化,否则判断的代码要复杂许多 总之就是边界都要添加” 哨兵” , 用于...
class Solution(object): def minPathSum(self, grid): """ :type grid: List[List[int]] :rtype: int """ row = len(grid)-1 column = len(grid[0])-1 i=row-1 j=column-1 while j>=0: grid[row][j]+=grid[row][j+1] j-=1 while i>=0: grid[i][column]+=grid[i+1][...