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.

Credits:
Special thanks to @porker2008 for adding this problem and creating all test cases.

给出乱序数组 要求排好序后相邻元素间的最大差值

由于要求是线性时间复杂度 所以所有内部排序方法都不可以用 只有用桶式排序（）参考http://blog.csdn.net/apei830/article/details/6596057

代码如下：

public class Solution {    public int maximumGap(int[] num) {        if(num.length<2)return 0;        int max=-1;        int min=Integer.MAX_VALUE;                int step=(max-min)/num.length+1;        for(int i=0;i<num.length;i++){            if(num[i]>max)max=num[i];            if(num[i]<min)min=num[i];        }        if(num.length==2)return max-min;        int[] minxx=new int[num.length+1];        int[] maxxx=new int[num.length+1];        for(int i=0;i<num.length;i++){            int x = num[i];            int k =  (int)(num.length * (1.0 * (x - min) / (max - min))); //attention! may have overflow problem!            if(minxx[k]==0 || x<minxx[k]) minxx[k] = x;            if(maxxx[k]==0 || x>maxxx[k]) maxxx[k] = x;        }        int maxInter = 0;        min = maxxx[0];        for(int i=1; i<num.length+1; i++) {            if(minxx[i] == 0) continue;            maxInter = Math.max(maxInter, minxx[i] - min);            min = maxxx[i];        }        return maxInter;    }}

Java-Maximum Gap