220. 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 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

难度：medium

题目：给定一整数数组，找出是否存在两个不同的索引i, j使其索引差的绝对值小于等于k, 值的差的绝对值小于等于t.

思路：

1. 暴力破解，
2. 使用滑动窗口和TreeSet是为了使得滑动窗口有序，TreeSet底层是红黑树, 如果暴力破解时间复杂度为O(kn), 改用TreeSet使得搜索时间复杂度为O(log K), 故总的时间复杂度为O(nlog K)。

Runtime: 22 ms, faster than 70.38% of Java online submissions for Contains Duplicate III.
Memory Usage: 40.4 MB, less than 5.58% of Java online submissions for Contains Duplicate III.

class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if(nums == null || nums.length < 2 || k < 1 || t < 0){
            return false;
        }
        TreeSet<Long> treeSet = new TreeSet<>();
        for (int i = 0; i < nums.length; i++) {
            if (!treeSet.subSet((long) nums[i] - t, true, (long) nums[i] + t, true).isEmpty()) {
                return true;
            }
            if (i >= k) {
                treeSet.remove((long) nums[i - k]);
            }
            treeSet.add((long) nums[i]);
        }
                    
        return false;
    }
}

Runtime: 987 ms, faster than 5.06% of Java online submissions for Contains Duplicate III.
Memory Usage: 38.8 MB, less than 56.75% of Java online submissions for Contains Duplicate III.

class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < Math.min(nums.length, i + k + 1); j++) {
                if (nums[i] > 0 && nums[j] < 0 || nums[i] < 0 && nums[j] > 0) {
                    if (Math.abs(nums[i]) - t > 0 
                        || Math.abs(nums[i]) - t + Math.abs(nums[j]) > 0) {
                        continue;
                    }
                }
                
                if (Math.abs(nums[i] - nums[j]) <= t) {
                    return true;
                }
            }
        }
                    
        return false;
    }
}