2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 publicclassSolution { publicbooleancontainsNearbyAlmostDuplicate(int[] nums,intk,intt) { if(k<1||t<0)returnfalse; TreeSet<Integer> set =newTreeSet<Integer>(); for(inti=0;i<nums.length;i++){ intn = nums[i]; if((set.floo...
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(...
publicbooleancontainsNearbyAlmostDuplicate(int[]nums,int k,int t){if(k<1||t<0||nums==null||nums.length<2){returnfalse;}SortedSet<Long>set=newTreeSet<Long>();for(int j=0;j<nums.length;j++){SortedSet<Long>subSet=set.subSet((long)nums[j]-t,(long)nums[j]+t+1);// 集合不为空,...
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])
示例 输入:nums = [1, 2, 3, 1], k = 3 输出:true 输入:nums = [1, ...219. Contains Duplicate II 题目: Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference ......
LeetCode 220. Contains Duplicate III 简介: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。 Description Given an array of integers, find out whether there are two distinct indices i ...
set.add(x); if(set.size()>k)set.remove(nums[i-k]); } //ensure return false; } } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 参考 [Leetcode] Contains Duplicate 包含重复...
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....
it));if (min_diff <= t)return true;}return false;}3 运行时间16ms，超过了百分之82%的程序。这个程序还是可以的。4 c#解法。public class Solution {public bool ContainsNearbyAlmostDuplicate(int[] nums, int k, int t) {var len = nums.Length;var arr = nums.Select(((num, index) => new ...
class Solution: def containsDuplicate(self, nums: List[int]) -> bool: # ## 解法三 s = set() for num in nums: if num in s: return True else: s.add(num) return False 解法四居然表现比解法三更高效。当然,他们的复杂度是一样的。 时间复杂度:O(n) 空间复杂度:O(n) (额外使用了一个...