220. Contains Duplicates

题目：
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：
这一题有两个思路，都是参考discussion里写出来的。一个是bucket, 一个是TreeSet。
1.bucket是按照两个数最多相差t这个性质，把每个数分到不一样的bucket里，在k范围内，如果有两个数在同一个bucket里，那么说明这两个数满足条件；或者相邻的bucket里存在一个数，而且与这个数的差小于等于t，那这个数也满足；其它都是超出范围的。

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来看有没有在这个范围内存在的数。

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;
}