leetcode 125 Valid Palindrome

最新推荐文章于 2020-05-27 12:11:01 发布

walter1990

最新推荐文章于 2020-05-27 12:11:01 发布

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本文介绍了一种算法，用于判断一个给定的字符串是否为回文，仅考虑字母数字字符并忽略大小写。通过两个指针从两端向中间扫描，跳过非字母数字字符进行比较。

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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.

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class Solution {
public:
	/*bool isPalindrome(string s) {
		int len = s.length();
		string str="";

		if(len==0) return true;

		for(int i = 0; i < len; i++) {
			if(s[i] >= 'a' && s[i] <= 'z' || s[i] >= '0' && s[i] <= '9') str += s[i];
			else if(s[i] >= 'A' && s[i] <= 'Z') str += (s[i]-'A'+'a');
		}

		for(int i = 0, j = len-1; i <= j; i++, j--) {
			if(str[i]!=str[j]) return false;
		}

		return true;
	}*/

	bool isPalindrome(string s) {
		int len = s.length();
		char pre, after;
		if(len == 0) return true;

		for(int i=0, j=len-1; i <= j;) {
			if(s[i] >= '0' && s[i] <= '9' || s[i] >= 'a' && s[i] <= 'z') {
				pre = s[i];
			} else if(s[i] >= 'A' && s[i] <= 'Z'){
				pre = s[i]-'A'+'a';
			} else {
				i++;
				continue;
			}

			if(s[j] >= '0' && s[j] <= '9' || s[j] >= 'a' && s[j] <= 'z') {
				after = s[j];
			} else if(s[j] >= 'A' && s[j] <= 'Z') {
				after = s[j]-'A'+'a';
			} else {
				j--
				continue;
			}
			i++;
			j--;
			if(after!=pre) return false;
		}

		return true;
	}
}