1 class Solution { 2 public int minSubArrayLen(int s, int[] nums) { 3 return sub(0,s,nums);//开始寻找 4 } 5 public int sub(int i,int s,int[] nums)//i为当前其实坐标,s和值,nums数组 6 { 7 if(i>=nums.length) return 0;//当达到数组尾
Given an array ofnpositive integers and a positive integers, find the minimal length of a subarray of which the sum ≥s. If there isn't one, return 0 instead. For example, given the array[2,3,1,2,4,3]ands = 7, the subarray[4,3]has the minimal length under the problem constraint...
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. For example, given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length ...
n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead. [2,3,1,2,4,3] and s = 7, the subarray [4,3] click to show more practice....
题意是:要求数组中的数字或数字之后大于或等于target, 求最小长度; 思路是:用双指针去探形成一个窗口,滑动窗口不仅只是窗口会滑动,窗口的大小也会变化;这里用一个for循环和一个while循环,for循环控制right, while循环控制left的变化对sum的影响,最后res每个阶段 或 res每个满足条件的阶段都已保存; ...
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...
Can you solve this real interview question? Sum of Subarray Minimums - Given an array of integers arr, find the sum of min(b), where b ranges over every (contiguous) subarray of arr. Since the answer may be large, return the answer modulo 109 + 7. Ex
position i and the value of the above parameter is p. For all subarrays [i, j] such that i <= j < p, the minimum will be arr[i]. And for all subarrays [i, j] such that p <= j <= n, the sum of minimum of these subarrays has already been computed and stored in DP[p...
(a % MOD * b % MOD) % MOD; } int sumSubarrayMins(vector<int>& A) { stack <int> st; int n = A.size(); vector <int> left(n, -1); vector <int> right(n, n); int ans = 0; for(int i = 0; i < n; i++){ while(!st.empty() && A[st.top()] >= A[i]){ ...
209. Minimum Size Subarray Sumwindliang 互联网行业 开发工程师 来自专栏 · LeetCode刷题 2 人赞同了该文章 题目描述(中等难度) 找出最小的连续子数组,使得子数组的和大于等于 s。 解法一 暴力破解 从第0 个数字开始,依次添加数字,记录当总和大于等于 s 时的长度。 从第1 个数字开始,依次添加...