不用比较两次,直接找nums[i] - t的lower_bound, 这个值就是与nums[i]差值最近的值。 1classSolution {2public:3boolcontainsNearbyAlmostDuplicate(vector<int>& nums,intk,intt) {4multiset<longlong>bst;5for(inti =0; i < nums.size(); ++i) {6if(bst.size() == k +1) bst.erase(bst.find(...
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] +...
代码(Python3) from sortedcontainers import SortedSet class Solution: def containsNearbyAlmostDuplicate(self, nums: List[int], index_diff: int, value_diff: int) -> bool: # num_set 维护滑动窗口 [i - index_diff, i) 内的所有数 num_set: SortedSet = SortedSet() for i, num in enumerate(...
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 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/...
【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. ...
Leetcode 220: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 absolute difference between nums[i] and nums[j] is at most t and the absolute difference between i and j is at most k....
LeetCode-220. Contains Duplicate III Given an array of integers, find out whether there are two distinct indicesiandjin the array such that theabsolutedifference betweennums[i]andnums[j]is at mosttand theabsolutedifference betweeniandjis at mostk....
专栏集合:王几行xing:【Python-转码刷题】LeetCode 力扣新手村100题汇总 +刷题顺序 读题 解法一:暴力解法 思路 逐一遍历元素; 将某个元素和它右边的元素逐一对比,如果相等则返回 True; 否则返回 False class Solution: def containsDuplicate(self, nums: List[int]) -> bool: for i in range(len(nums))...
Can you solve this real interview question? Contains Duplicate - Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct. Example 1: Input: nums = [1,2,3,1] Output: tr