Valid Palindrome

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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.

Two Pointers String

思路：1.回文只识别字母和数字；

2.大小写视为一样；

3.sting不是以'\0'结尾，所以用substr函数处理。

class Solution {public:        char lowerCase(char c){if (c >= 'A' && c <= 'Z')return c - 'A' + 'a';elsereturn c;}bool isPalindrome(string s) {int len = s.length();string temp(len,0);//因为空串不能用下标访问和赋值，所以得初始化int i = 0,j = 0;while (i < len){if (isalpha(s.at(i))){temp.at(j++) = s.at(i++);}else if (s.at(i) >= '0' && s.at(i) <= '9') {temp.at(j++) = s.at(i++);}else   i++;}//temp.at(j) = '\0';string类型不是以'\0'结尾temp = temp.substr(0, j);//从第0位取到第j位int len1 = temp.length();cout << len1 <<endl;for (int i = 0; i < len1/2; i++){if (lowerCase(temp[i]) == lowerCase(temp[len1 - 1 - i])){continue;}elsereturn false;}return true;}};

测试：27 ms

void main(){Solution s;string a;getline(cin, a);cout << s.isPalindrome(a);}

主要难点在于提供接口是string类型，需了解一些基本函数。

代码太冗余，经优化后：16 ms

                //判断是否为字母、数字//bool isalphanumeric(char c){函数执行完后，数组元素还是不会变为小写    bool isalphanumeric(char &c){if ((c >= 'A'&&c <= 'Z')){//大写则转换成小写c = c - 'A' + 'a';return true;}return (c >= '0'&&c <= '9') || (c >= 'a'&&c <= 'z');}        bool isPalindrome(string s) {int len = s.length();int i = 0, j = len - 1;while (i<j){if (!isalphanumeric(s[i])) {i++;continue;//跳出}if (!isalphanumeric(s[j])) {j--;continue;//跳出}if (s[i] != s[j]) return false;else              i++, j--;}return true;}

参考：http://blog.csdn.net/sunbaigui/article/details/8981357

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