220. Contains Duplicate III

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();
    }

    /**
     * 这道题目我思考了半天都没想出答案来，官方的答案又得订阅会员才能看到。。无奈之下看了下讨论区别人的答案，发现确实思路灰常不错，脑洞大开
     *
     * 妈的，刚开始以为自己看懂了呢。其实看懂了个屁。。这个方法：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;
    }

    // 然后，我又看了下提交区效率最高的答案，马丹，就是最简单的暴力方法
    // 奶奶的，不知道为啥我提交这个答案就是超时了。。。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;
    }
}