My code:

public class Solution {
    public int maximumGap(int[] nums) {
        if (nums == null || nums.length < 2) {
            return 0;
        }
        
        radixSort(nums);
        int max = Integer.MIN_VALUE;
        for (int i = 1; i < nums.length; i++) {
            max = Math.max(max, nums[i] - nums[i - 1]);
        }
        return max;
    }
    
    private void radixSort(int[] nums) {
        int max = getMax(nums);
        for (int digit = 1; digit <= max; digit *= 10) {
            countingSort(nums, digit);
        }
    }
    
    private int getMax(int[] nums) {
        int max = nums[0];
        for (int i = 1; i < nums.length; i++) {
            max = Math.max(max, nums[i]);
        }
        
        return max;
    }
    
    private void countingSort(int[] nums, int digit) {
        int[] ct = new int[10];
        for (int i = 0; i < nums.length; i++) {
            ct[(nums[i] / digit) % 10]++;
        }
        for (int i = 1; i < 10; i++) {
            ct[i] += ct[i - 1];
        }
        
        int[] temp = new int[nums.length];
        for (int i = nums.length - 1; i >= 0; i--) {
            temp[ct[(nums[i] / digit) % 10] - 1] = nums[i];
            ct[(nums[i] / digit) % 10]--;
        }
        for (int i = 0; i < nums.length; i++) {
            nums[i] = temp[i];
        }
    }
}

看到这道题目，第一个想法是排序。然后要求排序时间复杂度是 O(n)
就想到了 radix sort
但是忘记了是什么原理了，幸好以前总结了。温故了一下。

reference:
http://www.jianshu.com/p/d461a3f69499

code reference：
https://discuss.leetcode.com/topic/22221/radix-sort-solution-in-java-with-explanation/2

time complexity: O(n)
worst time complexity: O(10 * n) Integer.MAX_VALUE 一共十位
space complexity: O(n)

Anyway, Good luck, Richardo! — 09/28/2016