Leetcode: Contains Duplicate III

用BST来维护前k个数，因此对于每个数，删除第i-k-1个数是lgn，得到ceiling和lower key分别是lgn，最后将当前数加入BST是lgn。

public class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if (nums == null) {
            return false;
        }
        TreeMap<Long, Integer> map = new TreeMap<Long, Integer>();//pair: value-index
        for (int i = 0; i < nums.length; i++) {
            if (i > k) {
                map.remove((long)nums[i-k-1]);
            }
            long val = (long)nums[i];
            Long greater = map.ceilingKey(val);
            if (greater != null && greater - val <= t) {
                return true;
            }
            Long smaller = map.lowerKey(val);
            if (smaller != null && val - smaller <= t) {
                return true;
            }
            map.put(val, i);
        }
        return false;
    }
}