Microsoft笔试题--给定一维整数数组,找到数组里两两数之差绝对值最小的值

/*
This is a free Program, You can modify or redistribute it under the terms of GNU
*Description: 2011年Microsoft一道笔试题, 给定一维整数数组,找到数组里两两数之差绝对值最小的值
*Language: C++
*Development Environment: VC6.0
*Author: Wangzhicheng
*E-mail: [email protected]
*Date: 2013/3/16
*/
#include <iostream>
#include <cstdlib>
#include <vector>
#include <ctime>
#include <cmath>
#include <algorithm>
using namespace std;

class Solution {
private:
	int N;
	vector<int>array;
	int x, y;
	int min;
public:
	Solution(int n) {
		if(n <= 0) {
			perror("arguments error!\n");
		}
		N = n;
		int i;
		srand(unsigned(time(0)));
		for(i=0; i<n; i++) {
			int x = rand() % n;
			array.push_back(x);
			cout<<x<<" ";
		}
		cout<<endl;
		x = array[0];
		y = array[1];
		min = fabs(x-y);
	}
	/*
	 * 方法一: 采用枚举法, 时间复杂度为O(n*n)
	*/
	void JustDoIt1() {
		int i, j;
		for(i = 0; i< N - 1; i++) {
			for(j = i + 1; j < N; j++) {
				if(fabs(array[i] - array[j]) < min) {
					x = i;
					y = j;
					min = fabs(array[i] - array[j]);
				}
			}
		}
	}
	/*
	 * 方法二: 先快速排序, 后相邻两数两两比较, 时间复杂度为O(n*lg(n))
	*/
	void JustDoIt2() {
		vector<int>v;
		v.assign(array.begin(), array.end());
		sort(v.begin(), v.end());
		int x, y;
		int min;
		x = v[0];
		y = v[1];
		min = fabs(x-y);

		int i;
		for(i = 1; i< N; i++) {
			if(fabs(v[i] - v[i-1]) < min) {
				x = i;
				y = i - 1;
				min = fabs(v[i] - v[i-1]);
			}
		}
		cout<<v[x]<<"-"<<v[y]<<"的绝对值最小,是"<<min<<endl;
	}
	void show() const {
		cout<<array[x]<<"-"<<array[y]<<"的绝对值最小,是"<<min<<endl;
	}
};


void main() {
	Solution s(10);
	s.JustDoIt1();
	s.show();
	s.JustDoIt2();
}