【Leetcode】【Easy】Valid Palindrome

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.

解题思路：

建立两个index，同时从前从后遍历字符串。遇到非字符需要跳过，遇到大写字母将其转为小写字母。比较两个index指向的字母，如果一致则继续遍历下一组，如果不相同则返回false，最后返回true；

注意：

1、非字母不需要比较，要跳过；

2、注意统一大小写；

class Solution {
public:
    bool isPalindrome(string s) {
        int len = s.length();
        int front = 0;
        int behind = len - 1;
        
        if (!len) 
            return true;
            
        while (front <= behind) {
            
            if (!isAlphanumeric(&s[front])) {
                front++;
                continue;
            }
            
            if (!isAlphanumeric(&s[behind])) {
                behind--;
                continue;
            }

            if (s[front] != s[behind]) {
                return false;
            } else {
                front ++;
                behind --;
            }
                
        }
        
        return true;
    }
    
    bool isAlphanumeric(char *s) {
        if ((*s>='a' && *s<='z') || (*s>='0' && *s<='9'))
            return true;
        if (*s>='A' && *s<='Z') {
            *s += 32;
            return true;
        }
        return false;
    }
    
};

另，有些程序直接使用了C++中isalnum()和tolower()函数，思路是一样的：

class Solution {
public:
    bool isPalindrome(string s) {
        int len = s.length();
        int front = 0;
        int behind = len - 1;
        
        if (!len) 
            return true;
            
        while (front <= behind) {
            if (!isalnum(s[front])) {
                front++;
                continue;
            }
            
            if (!isalnum(s[behind])) {
                behind--;
                continue;
            }

            if (tolower(s[front]) != tolower(s[behind])) {
                return false;
            } else {
                front ++;
                behind --;
            }
        }
        
        return true;
    }
};

附录：

常用字符ASCII值