Java-Maximum Gap

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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;    }}

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