给定一个数组，求如果排序之后，相邻两数的最大差值，要求时
间复杂度O(N)，且要求不能用非基于比较的排序。（面试高频题）

代码如下：

public class MaxGap {
    public static int maxGap(int[] arr) {
        if (arr==null && arr.length<2) {
            return 0;
        }
        int len = arr.length;
        int mix = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for (int i=0; i<len; i++) {
            mix = Math.min(mix, arr[i]);
            max = Math.max(max, arr[i]);
        }
        if (mix == max) {
            return 0;
        }
        boolean[] hasnum = new boolean[len+1];
        int[] mixs = new int[len+1];
        int[] maxs = new int[len+1];
        for (int i=0; i<len; i++) {
            int bid = bucket(arr[i], len, mix, max);
            mixs[bid] = hasnum[i] ? Math.min(mixs[i], arr[i]):arr[i];
            maxs[bid] = hasnum[i] ? Math.max(maxs[i], arr[i]):arr[i];
            hasnum[bid] = true;
        }
        int res = 0;
        int lagest = maxs[0];
        int j=1;
        for (; j<len+1; j++) {
            if (hasnum[j]) {
                res = Math.max(res, mixs[j]-lagest);
                lagest = maxs[j];
            }
        }
        System.out.println(res);
        return res;
    }

    public static int bucket (long num, long len, long mix, long max) {
        return ((int)((num - mix)*len/(max-mix)));
    }

    public static void main(String[] args) {
        int[] arr = {2,3,5,1,9,8,4};
        //maxGap(arr);
        System.out.println(maxGap(arr));
    }
}