classSolution {public:boolcontainsNearbyAlmostDuplicate(vector<int>& nums,intk,intt) { multiset<int>set;for(inti=0;i<nums.size();i++) {if(set.size()==k+1) {set.erase(set.find(nums[i-k-1])); } auto it=set.lower_bound(nums[i]-t); //这儿不能用abs(nums[i]) 涉及到负数case...
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] + (long)t) {...
Remove(nums[i - indexDiff]) } } // 在循环中没有返回,则所有数都不满足题意 return false } 题目链接: Contains Duplicate III: leetcode.com/problems/c 存在重复元素 III: leetcode.cn/problems/co LeetCode 日更第 282 天,感谢阅读至此的你 欢迎点赞、收藏、在看鼓励支持小满...
= null && Math.abs((long) higherKey - nums[i]) <= t)) {return true;}}set.add(nums[i]);}return false;}6 python 解法。class Solution(object):def containsNearbyAlmostDuplicate(self, A, k, t):""":type A: List[int]:type k: int:type t: int:rtype: bool"""n = len(A)A = ...
题目地址:https://leetcode.com/problems/contains-duplicate-iii/description/ 思路:刚开始直接暴力,先是错误,再是超时 错误代码1:(这个是没有考虑num[i]-num[j]上溢的情况) publicclassSolution{ publicbooleancontainsNearbyAlmostDuplicate(int[]nums,intk,intt) { ...
1classSolution {2publicbooleancontainsNearbyAlmostDuplicate(int[] nums,intk,intt) {3//corner case4if(nums ==null|| nums.length < 2 || k == 0) {5returnfalse;6}78//normal case9TreeSet<Long> set =newTreeSet<>();10inti = 0;11while(i <nums.length) {12Long floor = set.floor((...
【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. ...
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])
var containsDuplicate = function(nums) { var hash = {} for(var i = 0; i < nums.length; i++){ var el = nums[i]; if(el in hash){ return true } hash[el] = true } return false }; 第218题,判定数组里面有重复项,并要求它们的索引值的绝对值不能大于 k, 因此我们需要将之前索引...
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