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;//当达到数组尾端返回0 8 int count=0,sum=0; 9 f...
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 npositiveintegers and a positive integer s, find theminimal lengthof acontiguous subarrayof which thesum ≥ s. If there isn’t one, return 0 instead. 说人话: 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,并返回其长...
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; ...
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, ...
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...
publicintminSubArrayLen(ints,int[]nums){intmin=Integer.MAX_VALUE;intn=nums.length;for(inti=0;i<n;i++){intstart=i;intsum=0;while(start<n){sum+=nums[start];start++;//当前和大于等于 s 的时候结束if(sum>=s){min=Math.min(min,start-i);break;}}}//min 是否更新,如果没有更新说明数...
Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A. Since the answer may be large, return the answer modulo 10^9 + 7. Example 1: Input: [3,1,2,4] Output: 17 Explanation: Subarrays are [3], [1], [2], [4], [3,...