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.

 

public class Solution {
    public int maximumGap(int[] nums) {
        if (nums == null || nums.length < 2) {
        	return 0;
        }
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < nums.length; i++) {
			min = Math.min(min, nums[i]);
			max = Math.max(max, nums[i]);
		}
        if (min == max) {
        	return 0;
        }
        boolean[] hasNum = new boolean[nums.length+1];
        int[] maxs = new int[nums.length+1];
        int[] mins = new int[nums.length+1];
        int bid = 0;
        for (int i = 0; i < nums.length; i++) {
			bid = bucket(nums[i], nums.length, min, max);
			mins[bid] = hasNum[bid] ? Math.min(mins[bid], nums[i]) : nums[i];
			maxs[bid] = hasNum[bid] ? Math.max(maxs[bid], nums[i]) : nums[i];
			hasNum[bid] = true;
		}
        int res = 0;
        int lastMax = 0;
        int i = 0;
        while (i <= nums.length) {
        	if (hasNum[i++]) {
        		lastMax = maxs[i-1];
        		break;
        	}
        }
        for (; i <= nums.length; i++) {
			if (hasNum[i]) {
				res = Math.max(res, mins[i]-lastMax);
				lastMax = maxs[i];
			}
		}
        return res;
    }

	private int bucket(long num, long length, long min, long max) {
		// TODO Auto-generated method stub
		return (int) ((num - min) * length / (max - min));
	}
}