每日微软面试题——day 6（打印所有对称子串）

<以下微软面试题全来自网络>

<以下答案与分析纯属个人观点，不足之处，还望不吝指出^_^>

<出处：http://blog.csdn.net/zhanxinhang>
题：1、如何判断一个字符串是对称的？如a，aa，aba。
2、如何利用2函数找出一个字符串中的所有对称子串？

分析：
且看第一个问题判断字符串对称，有以下两种方法。

方法一、使迭代器p1指向头，p2指向尾。使p1，p2相对而行(—> | <[b]—)[/b]，每次比较p1，p2是否相等，直到它们相遇。
方法二、使p1、p2指向中间的一个元素（如果字符串长度为偶数的话就指向中间两个相邻的元素），使它们相向而行(<[b]— |[b]—>[/b])[/b]，每次比较p1，p2是否相等。直到它们一个指向头一个指向尾便结束。

（可以看出方法1明显好于方法2）

现在看第二个问题打印所有的对称子串，判断对称子串的问题我们似乎已经解决，且有以上两种方法，针对现在的情况是否两种都合适呢，确切的说不是。如果是方法一，请想象一下，如果要p1，p2一个指向子串的头一个指向子串的尾，那么至少要两层循环一层控制p1一层控制p2，还有一层循环用于判断子串是否对称，也就是说总共需要三层嵌套循环，时间复杂度指数为3。而如果选择方法二呢？
两层循环就能实现，不过要适应现在这种情况需要做些修改，就是针对偶数长度的子串进行一次遍历，再针对奇数长度的子串进行一次遍历，也就是两层嵌套循环中内层有两次循环，时间复杂度指数为2。详情且看代码。

实现加测试代码：

/**
Author:花心龟
Blog：http://blog.csdn.net/zhanxinhang
**/

#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;
class App
{
	typedef string::iterator Iter_t;

	string m_str;
	void print(Iter_t p1, Iter_t p2) //打印从p1开始到p2结束的字符
	{
		while(p1<=p2)
		{
			cout<<*p1;
			p1++;
		}
		cout<<" | ";
	}
	bool isHuiwen1(Iter_t start,Iter_t end) //法1判断是否为对称子串
	{
		Iter_t p1 = start;
		Iter_t p2 = end;
		int times = (distance(p1,p2)+1) / 2; //比较次数
		while(times)
		{
			if(*p1 != *p2)
				return false;

			p1++; 
			p2--; 
			times--;
		}
		return true;
	}

	bool isHuiwen2(Iter_t mid_it) //法2判断是否为对称子串
	{
		Iter_t p1,p2;
		p1 = p2 = mid_it;

		while(p1>=m_str.begin() && p2 < m_str.end()) 
		{			
			if(*p1 != *p2) 
				break;	
			print(p1,p2);//打印从p1到p2的字符
			p1--,p2++;
		}

		p1 = p2 = mid_it;
		p2++; //使p2向前移动一个位置，此时p1，p2分别指向中间两个相邻位置
		while(p1>=m_str.begin() && p2 < m_str.end())  
		{
			if(*p1 != *p2)
				return false;
			print(p1,p2);//打印从p1到p2的字符
			p1--,p2++;
		}
		return true;
	}

	void show_all_substr_of_huiwen1() //法一打印所有对称子串
	{
		Iter_t p2=m_str.end()-1;
		Iter_t p1 = m_str.begin();
		for(;p1!=m_str.end();p1++)
			for(p2=p1;p2!=m_str.end();p2++)
			{
				if(isHuiwen1(p1,p2))
					print(p1,p2);
			}
	}

	void show_all_substr_of_huiwen2()  //法二打印所有对称子串
	{
		Iter_t it = m_str.begin();

		for(;it!=m_str.end();it++)
		{
			isHuiwen2(it);
		}
	}

public:

	void run()
	{
		m_str="abaaba";
		cout<<"测试数据为:abaaba"<<endl;
		show_all_substr_of_huiwen1();
		cout<<endl;
		show_all_substr_of_huiwen2();
		cout<<endl;

		m_str="asdfie";
		cout<<"测试数据为:asdfie"<<endl;
		show_all_substr_of_huiwen1();
		cout<<endl;
		show_all_substr_of_huiwen2();
		cout<<endl;

		m_str="aabaa";
		cout<<"测试数据为:aabaa"<<endl;
		show_all_substr_of_huiwen1();
		cout<<endl;
		show_all_substr_of_huiwen2();
		cout<<endl;

		//时间比较//
		m_str="this is a string for testing. aabaa alskjdfkljasdjflasdflkajsldkjfsjlakjsdlfjwoekjlakjlsdkjflsajlkdjfowieuoriuq aaddbb sldjfalkjsdlfkjasldjflajsldfjalskdjflakjsdlfkjaslkdjflajsdlfkjaslkdjflkajsdlkjflkasjdlfjaklsjdkfljaklsdjfklsajdflkjslkdjflaskjdlfkjalsdjlfkajsldfkjlaksjdfljasldjflaskjdfkasjdflaksjdkfljaskldfjlaksjdfljasldjflaksjdkljfkalsjdlkfjasldjflasjdlfjasldjfklsajdfljaskldfjlsakjdflkasjdfkl this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a string for testing.this is a 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		time_t start,record1;

		cout<<"show all substr of huiwen 1:";
		system("pause");
		start = time(NULL);    
		show_all_substr_of_huiwen1();
		record1 = time(NULL);

		cout<<endl;
		cout<<"time:"<<record1-start;
		cout<<endl;

		cout<<"show all substr of huiwen 2:";
		system("pause");
		start = time(NULL);
		show_all_substr_of_huiwen2();
		record1 = time(NULL);

		cout<<endl;
		cout<<"time:"<<record1-start;
		cout<<endl;

		cout<<"(可以看到打印速度明显快于上一次)";
	}
};

int main()
{
	App myapp;
	myapp.run();

	system("pause");
	return 0;
}

测试结果：



各测试数据下的一二行为打印出来的所有对称字串。
按下任意键后：



以上是使用方法1打印出来的结果。
按下任意键后：



以上便是用方法2打印出来的结果。

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