[Leetcode] Contains Duplicate III

最新推荐文章于 2020-04-25 23:41:16 发布

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最新推荐文章于 2020-04-25 23:41:16 发布

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原文链接：http://www.cnblogs.com/Phoebe815/p/4924766.html

本文介绍了一种算法问题的两种解决方案，旨在找出数组中是否存在两个不同的元素，这两个元素的值相差不超过t，且下标相差不超过k。第一种方法使用了TreeSet进行查找，第二种方法则采用了桶的概念来实现。

Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k.

SOL1:

 1 public class Solution {
 2     public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
 3         if(nums == null || nums.length == 0 || t < 0)
 4             return false;
 5         TreeSet<Long> ts = new TreeSet<>();
 6         for(int i = 0; i < nums.length; ++i) {
 7             if(i > k)
 8                 ts.remove((long)nums[i - k - 1]);
 9             TreeSet<Long> temp = (TreeSet<Long>) ts.subSet((long)nums[i] - t, true,(long)nums[i] + t, true);
10             if(!temp.isEmpty())
11                 return true;
12             ts.add((long)nums[i]);
13         }
14         return false;
15     }
16 }