[LeetCode-125] Valid Palindrome

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.

 

从外往里比对，跳过无效索引，不匹配返回false即可，注意处理空串以及全忽略的case。

 

class Solution {
public:
    //
    // 0 not valid
    // 1 valid
    // 2 valid, but is caps
    //
    inline int check_char(int idx, const string& s) {
        if ((s[idx] >= '0' && s[idx] <= '9') || (s[idx] >= 'a' && s[idx] <= 'z')) {
            return 1;
        } else if ((s[idx] >= 'A' && s[idx] <= 'Z')) {
            return 2;
        }
        return 0;
    }
    
    bool isPalindrome(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int idx_l = 0, idx_r = s.length() - 1, ret = 0;
        char cl = 0, cr = 0;
        while (idx_l < idx_r) {
            while (idx_l < s.length() && (0 == (ret = (check_char(idx_l, s))))) {
                ++idx_l;
            }
            if (idx_l < s.length()) {
                cl = (1 == ret ? s[idx_l] : s[idx_l] - 'A' + 'a');
            } else {
                cl = '\0';
            }
            while (idx_r >= 0 && (0 == (ret = (check_char(idx_r, s))))) {
                --idx_r;
            }
            if (idx_r >= 0) {
                cr = (1 == ret ? s[idx_r] : s[idx_r] - 'A' + 'a');
            } else {
                cr = '\0';
            }
            if ((idx_l < idx_r) && cl != cr) {
                return false;
            }
            ++idx_l;
            --idx_r;
        }
        return true;
    }
};