leetcode--Valid Palindrome

最新推荐文章于 2025-09-09 08:37:13 发布

转载最新推荐文章于 2025-09-09 08:37:13 发布 · 40 阅读

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原文链接：http://www.cnblogs.com/bingtel/p/4194424.html

文章标签：

#java

本文介绍如何使用Java和C++语言来验证一个字符串是否为回文，考虑仅包含字母数字字符并忽略大小写。文章还讨论了在面试中遇到的常见问题，并提供了代码实现。

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.

双指针问题：

java：

public class Solution {
    public boolean isPalindrome(String s) {
        int len = s.length();
        if(len<=0){
            return true;
        }
        int i=0,j=len-1;
        while(i<=j){
            if(s.charAt(i)>='a'&&s.charAt(i)<='z'||s.charAt(i)>='A'&&s.charAt(i)<='Z'||s.charAt(i)>='0'&&s.charAt(i)<='9'){
                if(s.charAt(j)>='a'&&s.charAt(j)<='z'||s.charAt(j)>='A'&&s.charAt(j)<='Z'||s.charAt(j)>='0'&&s.charAt(j)<='9'){
                    if(s.charAt(i)==s.charAt(j)||s.charAt(i)==s.charAt(j)+32||s.charAt(i)+32==s.charAt(j)){
                        i++;
                        j--;
                    }else{
                        return false;
                    }
                }else{
                    j--;
                }
            }else{
                i++;
            }
        }
        return true;
    }
}

c++：

class Solution {
public:
    bool isPalindrome(string s) {
        int len = s.length();
        if(len<=1)
            return true;
        for(int i=0;i<len;i++){
            if(s[i]>='a'&&s[i]<='z'){
                s[i]=(char)(s[i]-32);
            }
        }
        int i=0,j=len-1;
        while(i<=j){
            if(s[i]>='0'&&s[i]<='9'||s[i]>='A'&&s[i]<='Z'){
                if(s[j]>='0'&&s[j]<='9'||s[j]>='A'&&s[j]<='Z'){
                    if(s[i]!=s[j]){
                        return false;
                    }else{
                        j--;
                        i++;
                    }
                }else{
                    j--;
                }
            }else{
                i++;
            }
        }
        return true;
    }
};

转载于:https://www.cnblogs.com/bingtel/p/4194424.html