LeetCode刷题笔记（递归）：valid-palindrome

雨点打透心脏的1/2处 2022-06-01 09:41 89阅读 0赞

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*  **转载请注明作者和出处：[http://blog.csdn.net/u011475210][http_blog.csdn.net_u011475210]**
 *  **代码地址：[https://github.com/WordZzzz/Note/tree/master/LeetCode][https_github.com_WordZzzz_Note_tree_master_LeetCode]**
 *  **刷题平台：[https://www.nowcoder.com/ta/leetcode][https_www.nowcoder.com_ta_leetcode]**
 *  **题  库：Leetcode经典编程题**
 *  **编  者：WordZzzz**

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*   *  题目描述
     *  解题思路
     *  C版代码实现
        
         *  循环
         *  递归

## 题目描述 ##

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 ##

额，判断字符串是否为回文，这道题貌似没什么好说的，需要注意的是空字符串也算回文哦。

从两边开始遍历字符串，如果遇到非数字字符的就跳过，然后比较当前两个字符是否相等；相等就进入下一个循环，否则返回false。

## C++版代码实现 ##

### 循环 ###

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

### 递归 ###

class Solution {
    public:
    
        bool isPalindromeCore(string &s, int start, int end){
            if(start >= end)
                return true;
            while(start < end && !isalnum(s[start]))
                start++;
            while(start < end && !isalnum(s[end]))
                end--;
            if(start < end && tolower(s[start]) != tolower(s[end]))
                return false;
            return isPalindromeCore(s, ++start, --end);
        }
        bool isPalindrome(string s) {
            if(s.size() == 0)
                return true;
            return isPalindromeCore(s, 0, s.size());
        }
    };

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[http_blog.csdn.net_u011475210]: http://blog.csdn.net/u011475210
[https_github.com_WordZzzz_Note_tree_master_LeetCode]: https://github.com/WordZzzz/Note/tree/master/LeetCode
[https_www.nowcoder.com_ta_leetcode]: https://www.nowcoder.com/ta/leetcode