10 11 12 13 14 Given an array of integers, find out whether there are two distinct indices i and j in the array such that the absolute difference between nums[i] and nums[j] is at most t and the absolute difference between i and j is at most k. Example1: Input: nums = [1,2,...
3.1 Java实现 publicclassSolution{publicbooleancontainsNearbyAlmostDuplicate(int[] nums,intk,intt){intn=nums.length; TreeSet<Long> set =newTreeSet<>();for(inti=0; i < n; i++) {Longceiling=set.ceiling((long) nums[i] - (long) t);if(ceiling !=null&& ceiling <= (long) nums[i] +...
【摘要】 Leetcode 题目解析之 Contains Duplicate III Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between numsi and numsj is at most t and the difference between i and j is at most k. 对于加入的元素,要能够在...
import "github.com/emirpasic/gods/maps/treemap" func containsNearbyAlmostDuplicate(nums []int, indexDiff int, valueDiff int) bool { // numToIndex 维护滑动窗口 [i - indexDiff, i) 内的所有数 numToIndex := treemap.NewWithIntComparator() for i, num := range nums { // 如果滑动窗口内存...
Can you solve this real interview question? Contains Duplicate III - You are given an integer array nums and two integers indexDiff and valueDiff. Find a pair of indices (i, j) such that: * i != j, * abs(i - j) <= indexDiff. * abs(nums[i] - nums[j])
[LeetCode]Contains Duplicate III Question Given an array of integers, find out whether there are two distinct indices i and j in the array such that theabsolutedifference betweennums[i]andnums[j]is at most t and theabsolutedifference between i and j is at most k....
Output: false 1. 2. 题解: 使用一个容器存储每个数和它的下标。 然后排序,一是减小搜索量,二是防止溢出(set寻找一直爆)。 遍历数组找到第一个满足nums[i] + t >= nums[j]的,判断其下标是否在k内。 classSolution{ public: boolcontainsNearbyAlmostDuplicate(vector<int>&nums,intk,intt) { ...
Leetcode 220 Contains Duplicate III如何解决 简介 220存在重复元素。给定一个整数数组，判断数组中是否有两个不同的索引 i 和 j，使得nums [i] 和nums [j]的差的绝对值最大为 t，并且 i 和 j 之间的差的绝对值最大为 ķ。c++，python，c#，java等语言的解法 工具/原料 c++，python，c#，java 方法/...
class Solution {public: bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) { if(t<0) return false; //t+1可能会溢出，所以要+ 1LL long long mod = t + 1LL; unordered_map<long long,long long> buck; for(int i=0;i<nums.size();i++) { ...
classSolution{publicbooleancontainsNearbyAlmostDuplicate(int[]nums,int k,int t){int n=nums.length;TreeSet<Long>set=newTreeSet<Long>();for(int i=0;i<n;i++){Long ceiling=set.ceiling((long)nums[i]-(long)t);if(ceiling!=null&&ceiling<=(long)nums[i]+(long)t){returntrue;}set.add((...