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valid-palindrome——判断带符号数字字母的字符串是否为回文

2024-10-05 11:52:02   217人阅读

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.

 

字符串为空时判断为回文，大小写不区分可确定相等，数字与字母不同。

 

 1 class Solution {
 2 public:
 3     bool isPalindrome(string s) {
 4         int n=s.length();
 5         if(n==0) return true;
 6         int i=0,j=n-1;
 7         while(i<j){
 8             if(!isCharacters(s[i])){
 9                 i++;
10                 continue;
11             }
12             if(!isCharacters(s[j])){
13                 j--;
14                 continue;
15             }
16                 
17             int left=0,right=0,leftsig=0,rightsig=0;
18             left=(s[i]>=‘0‘&&s[i]<=‘9‘)?s[i]-‘0‘:((s[i]>=‘a‘&&s[i]<=‘z‘)?s[i]-‘a‘:s[i]-‘A‘);
19             right=(s[j]>=‘0‘&&s[j]<=‘9‘)?s[j]-‘0‘:((s[j]>=‘a‘&&s[j]<=‘z‘)?s[j]-‘a‘:s[j]-‘A‘);
20             leftsig=(s[i]>=‘0‘&&s[i]<=‘9‘)?0:1;
21             rightsig=(s[j]>=‘0‘&&s[j]<=‘9‘)?0:1;
22             if(left!=right||leftsig!=rightsig)
23                 return false;
24             i++;
25             j--;
26         }
27         return true;
28     }
29     bool isCharacters(char c){
30         if((c>=‘a‘&&c<=‘z‘)||(c>=‘A‘&&c<=‘Z‘)||(c>=‘0‘&&c<=‘9‘))
31             return true;
32         else
33             return false;
34     }
35 };

 

valid-palindrome——判断带符号数字字母的字符串是否为回文


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