[LeetCode]Maximum Gapd

一题一句：

int bucket_size = Math.max(1,(Max-Min)/(N) + 1);//ceil( (Max-Min)/(N-1) ) = (Max-Min-1)/(N-1) + 1        or int  bucket_size = (int)Math.ceil( ((double)(Max-Min))/(N-1) );	int bucket_count = (Max-Min)/bucket_size + 1;

下面用桶排序实现，这也是leetcode上给出的参考解法：

Suppose there are N elements and they range from A to B.

Then the maximum gap will be no smaller than ceiling[(B - A) / (N - 1)]

Let the length of a bucket to be len = ceiling[(B - A) / (N - 1)], then we will have at most num = (B - A) / len + 1 of bucket

for any number K in the array, we can easily find out which bucket it belongs by calculating loc = (K - A) / len and therefore maintain the maximum and minimum elements in each bucket.

Since the maximum difference between elements in the same buckets will be at most len - 1, so the final answer will not be taken from two elements in the same buckets.

For each non-empty buckets p, find the next non-empty buckets q, then q.min - p.max could be the potential answer to the question. Return the maximum of all those values.

补充：
Java array is a pain in ass especially after you have tasted python list. But still, need to get used to it.