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leetCode(51):Valid Palindrome

时间:2015-07-25 13:54:03      阅读:101      评论:0      收藏:0      [点我收藏+]

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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.

//用两个指针实现,前提是要忽略其他符号

class Solution {
public:
    bool isPalindrome(string s) {
        if(s.empty())
            return true;
        int i=0;
        int j=s.size()-1;
        for(;i<j;i++)
        {//i后移
            if((s[i]>='a' && s[i]<='z') || (s[i]>='A' && s[i]<='Z') || (s[i]>='0' && s[i]<='9'))
            {//找到第一个为字符或数字的字符
                for(;j>i;--j)
                {//j前移
                    if((s[j]>='a' && s[j]<='z') || (s[j]>='A' && s[j]<='Z') || (s[j]>='0' && s[j]<='9'))
                    {//找到第一个为字符或数字的字符
                        if(s[i]==s[j] || s[i]+'A'-'a'==s[j] || s[i]+'a'-'A'==s[j])
                        {//找到后时行判断
                            j--;//前进一步
                            break;
                        }
                        else
                            return false;
                    }
                }
            }
        }
        return true;
    }
};


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leetCode(51):Valid Palindrome

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原文地址:http://blog.csdn.net/walker19900515/article/details/47055147

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