Leetcode 220. Contains Duplicate III

Question:

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.

最開始考慮雙層循環遍歷，可是代碼提交上去提示超時，想一想也是，雙層循環時間複雜度是O(n^2)。後考慮使用TreeSet，時間複雜度爲O(nlogk).

Code: 

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

提交上去後顯示執行時間爲51ms. 不是很理想，查看最優的代碼，以下（執行時間是28ms，順便也學習下）

public class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if (nums == null || nums.length == 0 || t < 0 ||  k < 0) return false;
        if (k == 0) return false;

        Map<Long, Long> map = new HashMap<>();
        long w = (long) t + 1;
        for (int i = 0; i < nums.length; i++) {
            long bucket = getBucket(nums[i], w);
            if (map.containsKey(bucket)) {
                return true;
            } 
            if (map.containsKey(bucket - 1) && (Math.abs(map.get(bucket-1) - (long) nums[i]) <= t)) {
                return true;
            } 
            if (map.containsKey(bucket + 1) && (Math.abs(map.get(bucket+1) - (long) nums[i]) <= t)) {
                return true;
            }
            map.put(bucket, (long) nums[i]);
            if (i >= k) {
                map.remove(getBucket(nums[i-k], w));
            }
        }
        
        return false;
    }
    
    private long getBucket(long num, long w) {
        return (long) (num < 0 ? (num + 1)/w - 1 : num / w);
    }
}