leetcode 125. Valid Palindrome 回文数的判断

ゝ一纸荒年。 2022-06-08 03:17 144阅读 0赞

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.

这道题考查的是指定字符的回文数字的判断。

建议和[leetcode 680. Valid Palindrome II 去除一个字符的回文字符串判断 + 双指针][leetcode 680. Valid Palindrome II _ _] 一起学习

代码如下：

public class Solution 
    {
        
        //判断是否是回文数：这个题只考虑小写字符和数字，别的是不考虑的
        public boolean isPalindrome(String s) 
        {
            if(s==null || s.length()<=1)
                return true;
    
            s=s.replace(" ", "");
            s=s.toLowerCase();
            int i=0,j=s.length()-1;
            while(i<j)
            {
                if(isVaild(s.charAt(i))==false)
                {
                    i++;
                    continue;
                }else if(isVaild(s.charAt(j))==false)
                {
                    j--;
                    continue;
                }else if(s.charAt(i)!=s.charAt(j))
                        return false;
                else
                {
                        i++;
                        j--;
                }
            }
            return true;
        }
    
        //判断是否是我们要检查的字符
        boolean isVaild(char c)
        {
            if(c>='a' && c<='z' || c>='0' && c<='9')
                return true;
            else
                return false;
        }
    }

下面是C++的做法，就是做一个遍历，这道题最大的收获就是学会使用transform函数

代码如下：

#include <iostream>
    #include <vector>
    #include <algorithm>
    #include <climits>
    
    using namespace std;
    
    class Solution 
    {
    public:
        bool isPalindrome(string s)
        {
            string a = s;
            transform(s.begin(),s.end(),a.begin(),::tolower);
            int i = 0, j = a.length() - 1;
            while (i < j)
            {
                if (isVaild(a[i]) && isVaild(a[j]))
                {
                    if (a[i] != a[j])
                        return false;
                    else
                    {
                        i++;
                        j--;
                    }
                }
                else if(isVaild(a[i]) && !isVaild(a[j]))
                    j--;
                else if (!isVaild(a[i]) && isVaild(a[j]))
                    i++;
                else
                {
                    i++;
                    j--;
                }
            }
            return true;
        }
    
        bool isVaild(char a)
        {
            if (a >= '0' && a <= '9' || a >= 'a' && a <= 'z')
                return true;
            else
                return false;
        }
    };

[leetcode 680. Valid Palindrome II _ _]: http://blog.csdn.net/jackzhang_123/article/details/78880022