Java-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.

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;
    }
}