无序数组排序后相邻两数的最大差值（不能用计数排序）

给你ｎ个任意整数，求排序后相邻两个数之间的最大差值，这里n可能有10^5，整数为任意32位整型。要求求解算法的时间复杂度为O(n)。
思路描述参考：求无序数组排序后相邻俩数最大差值
代码C++描述如下：

#include <iostream>
using namespace std;
int max(int x, int y) 
{
	int z = x > y ? x : y;
	return z;
}
int min(int x, int y)
{
	int z = x > y ? y : x;
	return z;
}
int bucket(long num, long len, long min, long max)
{
	return (int)((num - min) * len / (max - min));
}
int maxGap(int nums[], int length) 
{
	if(nums == NULL || length < 2) {
		return 0;
	}	
	int minValue = 32768;
	int maxValue = -32767;
	for(int i = 0; i < length; i++) {
		minValue = min(minValue, nums[i]);
		maxValue = max(maxValue, nums[i]);
	}
	if(minValue == maxValue) {
		return 0;
	}
	int hasNum[length + 1] = {0};
	int maxs[length + 1] = {0};
	int mins[length + 1] = {0};
	int bid = 0;
	for(int i = 0; i < length; i++) {
		bid = bucket(nums[i], length, minValue, maxValue);
		mins[bid] = hasNum[bid] ? min(mins[bid], nums[i]) : nums[i];
		maxs[bid] = hasNum[bid] ? max(maxs[bid], nums[i]) : nums[i];
		hasNum[bid] = true;
	}
	int res = 0;
	int lastMax = maxs[0];
	int i = 1;
	for(; i <= length; i++) {
		if(hasNum[i]) {
			res = max(res, mins[i] - lastMax);
			lastMax = maxs[i];
		}	
	}
	return res;
} 
int main(){
	int nums[]={0,89,34,22};
	int length = sizeof (nums)/sizeof(nums[0]);
	int res = maxGap(nums,length);
	printf("res is %d\n",res);
}