LeetCode 220. Contains Duplicate III（检查重复）

原题网址：https://leetcode.com/problems/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 nums[i] and nums[j] is
at most t and
the difference between i and j is
at most k.

方法一：使用有序集合。

public class Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if (nums == null || nums.length <= 1 || k <= 0 || t < 0) return false;
if (t < 0) t = -t;
TreeSet<Long> ts = new TreeSet<>();
for(int i=0; i<nums.length; i++) {
if (i>k) ts.remove((long)nums[i-k-1]);
if (!ts.subSet((long)nums[i]-t, true, (long)nums[i]+t, true).isEmpty()) return true;
ts.add((long)nums[i]);
}
return false;
}
}

方法二：使用桶。

public class Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if (nums == null || nums.length <= 1 || k < 1 || t < 0) return false;
long min = nums[0];
for(int num: nums) min = Math.min(min, num);
Map<Long, Long> buckets = new HashMap<>();
for(int i=0; i<nums.length; i++) {
if (i>k) {
long key = (nums[i-k-1]-min)/((long)t+1);
buckets.remove(key);
}
long key = (nums[i]-min)/((long)t+1);
if (buckets.containsKey(key)) return true;
Long small = buckets.get(key-1);
if (small != null && nums[i]-small<=t) return true;
Long great = buckets.get(key+1);
if (great != null && great-nums[i]<=t) return true;
buckets.put(key, (long)nums[i]);
}
return false;
}
}

心得：凡是涉及到间隔、距离的，都有可能应用桶排序。而且桶排序不一定用数组！