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.

一刷
利用bucket, bucket中最大最小值的差范围在t+1之间，于是要寻找的值只可能出现在相同的bucket或相邻的bucket。

1. 遍历数组中的元素，i表示index, 如果index>=k, 那么要把之前存在map中，index小于i-k的元素删除。
2. Bucket，一个bucket中的两个元素的value差小于等于t
   //[0, t], [t+1, 2t+1], [2t+2, 3t+2]

每次遍历一个元素，

1. 如果该bucket index已经存在map的key中，return true;
2. 如果bucket index-1已经存在map的key中, 且value与nums[i]的value绝对值的差小于w, return true;
3. 如果bucket index+1已经存在map的key中, 且value与nums[i]的value绝对值的差小于w, return true;

public class Solution {
        private long getID(long i, long w) {
        //[0, t], [t+1, 2t+1], [2t+2, 3t+2]
        return i < 0 ? (i + 1) / w - 1 : i / w;
    }

    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if (t < 0) return false;
        Map d = new HashMap<>();
        long w = (long)t + 1;
        for (int i = 0; i < nums.length; ++i) {
            long m = getID(nums[i], w);
            if (d.containsKey(m))
                return true;
            if (d.containsKey(m - 1) && Math.abs(nums[i] - d.get(m - 1)) < w)
                return true;
            if (d.containsKey(m + 1) && Math.abs(nums[i] - d.get(m + 1)) < w)
                return true;
            d.put(m, (long)nums[i]);
            if (i >= k) d.remove(getID(nums[i - k], w));
        }
        return false;
    }
}