原題鏈接：https://leetcode.com/problems/contains-duplicate-iii/description/
這道題目不會做，別人的答案也沒看懂。以後有錢了訂閱會員看官方答案吧：

import java.util.HashMap;
import java.util.Map;

/**
 * Created by clearbug on 2018/2/26.
 */
public class Solution {

    public static void main(String[] args) {
        Solution s = new Solution();
    }

    /**
 

     * 這道題目我思考了半天都沒想出答案來，官方的答案又得訂閱會員才能看到。。無奈之下看了下討論區別人的答案，發現確實思路灰常不錯，腦洞大開
     *
     * 媽的，剛開始以為自己看懂了呢。其實看懂了個屁。。這個方法：beats 91.94 %
     *
     * @param nums
     * @param k
     * @param t
     * @return
     */
    public boolean containsNearbyAlmostDuplicate1(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;
    }

    // 然後，我又看了下提交區效率最高的答案，馬丹，就是最簡單的暴力方法
    // 奶奶的，不知道為啥我提交這個答案就是超時了。。。不玩了
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if (k < 1 || t < 0 || nums.length <= 1) {
            return false;
        }

        for (int i = 0; i < nums.length - 1; i++) {
            for (int j = i + 1; j < nums.length && (j - i) <= k; j++) {
                if (Math.abs((long) nums[j] - nums[i]) <= t) {
                    return true;
                }
            }
        }

        return false;
    }
}

220. Contains Duplicate III