Leetcode: Maximum Gap

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Try to solve it in linear time/space.

Return 0 if the array contains less than 2 elements.

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

官方版（桶排序）：

假设有N个元素A到B。

那么最大差值不会小于ceiling[(B - A) / (N - 1)]

令bucket（桶）的大小len = ceiling[(B - A) / (N - 1)]

对于数组中的任意整数K，很容易通过算式loc = (K - A) / len找出其桶的位置，然后维护每一个桶的最大值和最小值

由于同一个桶内的元素之间的差值至多为len - 1，因此最终答案不会从同一个桶中选择。

对于每一个非空的桶p，找出下一个非空的桶q，则q.min - p.max可能就是备选答案。返回所有这些可能值中的最大值。

这里A是min，B是max，桶有num.length - 1个。min, max不参与放入桶中，除了min和max之外还有N-2个数字和N-1个桶，所以一定有一个空桶。因为有空桶的存在所以要用一个previous变量来代表上一个非空的桶的max。previous初始化为min，这样min就考虑了虽然min不在桶中。还要记得考虑max，所以最后遍历了桶之后还要再比一次max

本算法O(n)时间和空间

还要注意math.floor,不能用，要用math.ceil比如：2/2.667 = 0.7499062617172854, 期望是：1，但是floor会给0，ceil才能给1.

public class Solution {
    public int maximumGap(int[] num) {
        if (num == null || num.length < 2)
            return 0;
        // get the max and min value of the array
        int min = num[0];
        int max = num[0];
        for (int i:num) {
            min = Math.min(min, i);
            max = Math.max(max, i);
        }
        // the minimum possibale gap, ceiling of the integer division
        int gap = (int)Math.ceil((double)(max - min)/(num.length - 1));
        int[] bucketsMIN = new int[num.length - 1]; // store the min value in that bucket
        int[] bucketsMAX = new int[num.length - 1]; // store the max value in that bucket
        Arrays.fill(bucketsMIN, Integer.MAX_VALUE);
        Arrays.fill(bucketsMAX, Integer.MIN_VALUE);
        // put numbers into buckets
        for (int i:num) {
            if (i == min || i == max)
                continue;
            int idx = (i - min) / gap; // index of the right position in the buckets
            bucketsMIN[idx] = Math.min(i, bucketsMIN[idx]);
            bucketsMAX[idx] = Math.max(i, bucketsMAX[idx]);
        }
        // scan the buckets for the max gap
        int maxGap = Integer.MIN_VALUE;
        int previous = min;
        for (int i = 0; i < num.length - 1; i++) {
            if (bucketsMIN[i] == Integer.MAX_VALUE && bucketsMAX[i] == Integer.MIN_VALUE)
                // empty bucket
                continue;
            // min value minus the previous value is the current gap
            maxGap = Math.max(maxGap, bucketsMIN[i] - previous);
            // update previous bucket value
            previous = bucketsMAX[i];
        }
        maxGap = Math.max(maxGap, max - previous); // updata the final max value gap
        return maxGap;
    }
}