#include <iostream>#include <stdio.h>#include <vector>#include <algorithm>#include <string>using namespace std;/*問題:Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.For example,"A man, a plan, a canal: Panama" is a palindrome."race a car" is not a palindrome.Note:Have you consider that the string might be empty? This is a good question to ask during an interview.For the purpose of this PRoblem, we define empty string as valid palindrome.分析:此題實際是判斷一個字符串是否是回文字符串，不考慮大小寫，比較的時候只比較字母和數字，對于其他字符默認忽略。設定從前向后和從后向前兩個指針分別為low ,highA[low]或者A[high]不是數字或字母，則直接過濾，否則統一將字符轉換為小寫，并比較是否相等low >= high的時候結束易錯：如果是空字符串，直接是回文的。容易遺漏輸入:A man, a plan, a canal: Panamarace a car輸出：truefalse關鍵:1 易錯：如果是空字符串，直接是回文的。容易遺漏2 將字符串轉換為小寫，用的是transform函數(beg , end , where  , func)transform(s.begin() , s.end() , s.begin() , tolower);*/class Solution {public:    bool isPalindrome(string s) {		if(s.empty())      		{			return true;		}		//將字符串轉換為小寫，用的是transform函數(beg , end , where  , func)		//transform(s.begin() , s.end() , s.begin() , tolower);		int len = s.length();		for(int i = 0 ; i < len ; i++)		{			if('A' <= s.at(i) && s.at(i) <= 'Z')			{				s.at(i) = s.at(i) - 'A' + 'a';			}		}		int low = 0;		int high = len - 1;		while(low < high)		{			while(low < high && !(				 ('0' <= s.at(low) && s.at(low) <= '9') || 				 ('a' <= s.at(low) && s.at(low) <= 'z') ) )			{				low++;			}			if(low >= high)			{				return true;			}			while(low < high && !(				 ('0' <= s.at(high) && s.at(high) <= '9') || 				 ('a' <= s.at(high) && s.at(high) <= 'z') ) )			{				high--;			}			if(low >= high)			{				return true;			}			//比較兩個字符是否相同,先轉換為小寫,比較之后，還需要變換下標			if(s.at(low) != s.at(high))			{				return false;			}			low++;			high--;		}		return true;    }};void process(){	 vector<int> nums;	 Solution solution;	 vector<int> result;	 char str[1024];	 //對于有空格的字符串，直接用gets	 while(gets(str) )	 {		 string value(str);		 bool isOk = solution.isPalindrome(value);		 if(isOk)		 {			cout << "true" << endl;		 }		 else		 {			 cout << "false" << endl;		 }	 }}int main(int argc , char* argv[]){	process();	getchar();	return 0;}