Contains Duplicate III

本博文参考自：https://leetcode.com/discuss/38206/ac-o-n-solution-in-java-using-buckets-with-explanation

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.

判断在一个数组中，是否存在两个元素nums[i]和nums[j]，满足nums[i]和nums[j]之间的最大差为t，i和j之间最大的差为k。

这道题目，最直观的方法就是直接做，时间复杂度为O（n*k）

一种可行的解法是使用筒排序

程序如下：

public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if(k<1||t<0)
return false;
Map<Long, Long> map=new HashMap<Long, Long>();
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;
}