LeetCode Valid Palindrome
2015-05-15 10:47
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题目
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
For example,
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.
判断是否是回文,比较常规和简单的问题。其中需要跳过不是字母和数字的内容。
空字符串也属于回文。
两个指针分别从头和尾两端向中间扫描,跳过无效字符,判断是否相等即可。
代码:
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.
判断是否是回文,比较常规和简单的问题。其中需要跳过不是字母和数字的内容。
空字符串也属于回文。
两个指针分别从头和尾两端向中间扫描,跳过无效字符,判断是否相等即可。
代码:
class Solution { public: bool isPalindrome(string s) { int left=0,right=s.size()-1; //两头向中间扫描 while(left<right) { while(left<s.size()&&!isalnum(s[left])) //跳过无效字符 left++; while(right>=0&&!isalnum(s[right])) right--; if(left>=right) //结束 return true; if(s[left]>='A'&&s[left]<='Z') //转为小写 s[left]+='a'-'A'; if(s[right]>='A'&&s[right]<='Z') s[right]+='a'-'A'; if(s[left]!=s[right]) //不等 return false; else //相等 { left++; right--; } } return true; } };
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