220. Contains Duplicates

題目：
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.

解答：
這一題有兩個思路，都是參考discussion裏寫出來的。一個是bucket, 一個是TreeSet。
1.bucket是按照兩個數最多相差t這個性質，把每一個數分到不同的bucket裏，在k範圍內，若是有兩個數在同一個bucket裏，那麼說明這兩個數知足條件；或者相鄰的bucket裏存在一個數，並且與這個數的差小於等於t，那這個數也知足；其它都是超出範圍的。

public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
    if (k < 1 || t < 0) return false;
    Map<Long, Long> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        long remappedNum = (long) nums[i] - Integer.MIN_VALUE;
        long bucket = remappedNum / ((long) t + 1);
        if (map.containsKey(bucket) || (map.containsKey(bucket - 1) && remappedNum - map.get(bucket - 1) <= t) || (map.containsKey(bucket + 1) && map.get(bucket + 1) - remappedNum <= t)) {
            return true;
        }
        if (map.entrySet().size() >= k) {
            long lastBucket = ((long) nums[i - k] - Integer.MIN_VALUE) / ((long) t + 1);
            map.remove(lastBucket);
        }
        map.put(bucket, remappedNum);
    }
    return false;
}

2.TreeSet是更加直接的用floor和ceiling來看有沒有在這個範圍內存在的數。

public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
    if (k < 1 || t < 0) return false;
    TreeSet<Integer> values = new TreeSet<Integer>(); 
    for (int i = 0; i < nums.length; i++) {
        Integer floor = values.floor(nums[i] + t);
        Integer ceiling = values.ceiling(nums[i] - t);
        if ((floor != null && floor >= nums[i]) || (ceiling != null && ceiling <= nums[i])) {
            return true;
        }
        if (values.size() >= k) {
            values.remove(nums[i - k]);
        }
        values.add(nums[i]);
    }
    return false;
}