[LeetCode] 125. Valid Palindrome 验证回文字符串

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.

验证一个给定的字符串是否为回文，用两个指针分别指向字符的首尾，判断是否相同，相同就的都向中间移动1位，判断下一组，左指针小于右指针就一直循环。遇到标点符号就跳过，处理下一个。遇到大写字母就转换成小写字母。

Java：

class Solution {
    public boolean isPalindrome(String s) {
        char[] chs = s.toCharArray();
        int left = 0, right = s.length() - 1;
        while (left <= right) {
            while (left < right && !Character.isLetterOrDigit(chs[left]))
                left++;
            while (left < right && !Character.isLetterOrDigit(chs[right]))
                right--;
            if (Character.toLowerCase(chs[left++]) != Character.toLowerCase(chs[right--]))
                return false;
        }
        return true;
    }
}

Java:

public class Solution {
    public boolean isPalindrome(String s) {
        int size = s.length(), i = 0, j = size - 1;
        s = s.toLowerCase();

        while (i < j) {
            if (!(s.charAt(i) >= 'a' && s.charAt(i) <= 'z') && !(s.charAt(i) >= '0' && s.charAt(i) <= '9')) {
                i++;
            }
            else if (!(s.charAt(j) >= 'a' && s.charAt(j) <= 'z') && !(s.charAt(j) >= '0' && s.charAt(j) <= '9')) {
                j--;
            }
            else {
                if (s.charAt(i) != s.charAt(j)) return false;

                i++;
                j--;
            }
        }
        return true;
    }
}

Python：

class Solution:
    def isPalindrome(self, s):
        i, j = 0, len(s) - 1
        while i < j:
            while i < j and not s[i].isalnum():
                i += 1
            while i < j and not s[j].isalnum():
                j -= 1
            if s[i].lower() != s[j].lower():
                return False
            i, j = i + 1, j - 1
        return True

C++：

class Solution {
public:
    bool isPalindrome(string s) {
        int left = 0, right = s.size() - 1 ;
        while (left < right) {
            if (!isalnum(s[left])) ++left;
            else if (!isalnum(s[right])) --right;
            else if ((s[left] + 32 - 'a') %32 != (s[right] + 32 - 'a') % 32) return false;
            else {
                ++left; --right;
            }
        }
        return true;
    }
};

C++：

class Solution {
public:
    bool isPalindrome(string s) {
        int l = 0, r = s.size() - 1;
        while(l <= r){
            while(!isalnum(s[l]) && l < r) l++;
            while(!isalnum(s[r]) && l < r) r--;
            if(toupper(s[l]) != toupper(s[r])) return false;
            l++, r--;
        }
        return true;
    }
};

类似题目：

[LeetCode] 9. Palindrome Number 验证回文数字

[LeetCode] 5. Longest Palindromic Substring 最长回文子串

[LeetCode] 516. Longest Palindromic Subsequence 最长回文子序列

　　

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