res = sum(A[i] * f(i)) where f(i) is the number of subarrays, in which A[i] is the minimum. 难点就在于求f(i),为了求f(i)需要求left[i]和right[i]。 left[i]:A[i]左边严格大于A[i]的个数 right[i]:A[i]右边大于等于A[i]的个数 f(i) = (left[i] + 1) * (right[i...
publicclassSolution {publicintminSubArrayLen(ints,int[] nums) {if(nums ==null|| nums.length == 0)return0;intleft = 0, right = 0, res = nums.length + 1;intsum = 0;while(right <nums.length) {while(sum < s && right <nums.length) { sum+= nums[right++]; }while(sum >=s) ...
AI检测代码解析 packageleetcodefuncminSubArrayLen(targetint,nums[]int)int{left,sum,res:=0,0,len(nums)+1forright,v:=rangenums{sum+=vforsum>=target{res=min(res,right-left+1)sum-=nums[left]left++}}ifres==len(nums)+1{return0}returnres}funcmin(aint,bint)int{ifa>b{returnb}returna} ...
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)。 AI检测代码解析 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(...
第一种是prefix sum + 双指针。 例如Minimum Size Subarray Sum - LeetCode: Given an array of n positive integers and a positive integers, find the minimal length of a contiguous subarray of which the sum ≥s. If there isn't one, return 0 instead. 用O(N)时间得到一个prefix sum array,可...
2.Smallest Subarray with a given sum(easy) 2.1 问题描述 2.2 解决方法 2.3 代码 3.课后回顾 4.参考链接 sliding window pattern 1.原理描述 滑动窗口模式(sliding window pattern)是用于在给定数组或链表的特定窗口大小上执行所需的操作,比如寻找包含所有 1 的最长子数组。从第一个元素开始滑动窗口并逐个元素...
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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. ...
首先从起点j开始加,一旦超过sum, 则从头部开始减去已经加上的。直至小于sum, 在这个过程中不停地update min_len publicclassSolution{publicintminSubArrayLen(int s,int[]nums){if(nums==null||nums.length==0)return0;int i=0,j=0,sum=0,min=Integer.MAX_VALUE;while(j<nums.length){sum+=nums[j];...
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