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# 220.Contains-Duplicate-III

## 题目描述

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.
Example 1:
Input: nums = [1,2,3,1], k = 3, t = 0
Output: true
Example 2:
Input: nums = [1,0,1,1], k = 1, t = 2
Output: true
Example 3:
Input: nums = [1,5,9,1,5,9], k = 2, t = 3
Output: false

## 代码

### Approach #1 TreeSet - ceiling & floor

class Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
TreeSet<Integer> set = new TreeSet();
for (int i = 0; i < nums.length; i++) {
// Find the successor of current element
Integer s = set.ceiling(nums[i]);
if (s != null && s <= nums[i] + t) return true;
// Find the predecessor of current element
Integer g = set.floor(nums[i]);
if (g != null && nums[i] <= g + t) return true;
if (set.size() > k) {
set.remove(nums[i - k]);
}
}
return false;
}
}

### Approach #2 Buckets

class Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if (t < 0) return false;
Map<Long, Long> d = new HashMap();
long w = (long)t + 1;
for (int i = 0; i < nums.length; i++) {
long m = getId(nums[i], w);
if (d.containsKey(m)) return true;
if (d.containsKey(m - 1) && Math.abs(nums[i] - d.get(m - 1)) < w) {
return true;
}
if (d.containsKey(m + 1) && Math.abs(nums[i] - d.get(m + 1)) < w) {
return true;
}
d.put(m, (long)nums[i]);
}
return false;
}
private long getID(long x, long w) {
return x < 0 ? (x + 1) / w - 1 : x / w;
}
}