Leetcode 0220: 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.

中文描述：

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

Constraints:

0 <= nums.length <= 2 * 10⁴
-2³¹ <= nums[i] <= 2³¹ - 1
0 <= k <= 10⁴
0 <= t <= 2³¹ - 1

Time complexity: O(nlogk)
滑动窗口+Binary Search Tree：

1. 维持一个区间为k的滑动窗口，这样可以保证滑动窗口内的元素的差的绝对值小于等于k
2. 在滑动窗口中寻找与nums[i]最接近的数，即找滑动窗口中小于等于nums[i]的最大值，找滑动窗口中大于等于nums[i]的最小值，若他们任何一个与nums[i]的差的绝对值小于等于t，则直接返回true。注意int 的范围，以及要用TreeSet来维护滑动窗口，因为floor(E x)可以找到小于等于e的最大值，ceiling(E x)可以找到大于等于e的最小值，并且TreeSet本身是排序好的
   3、枚举完i位置后，需要将nums[i - k]位置的元素移除滑动窗口，将num[i]位置的元素加入到窗口

class Solution {
    
    
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
    
    
        TreeSet<Long> set = new TreeSet<>();
        for(int i = 0; i < nums.length; i++){
    
    
            Long n = (long)nums[i];
            Long l = set.floor(n);
            Long r = set.ceiling(n);
            if (l != null && n - l <= t) return true;
            if (r != null && r - n <= t) return true;
            set.add(n);
            if (set.size() > k) {
    
    
                set.remove((long)nums[i - k]);
            }
        }
        return false;
    }
}