[LeetCode] 125. Valid Palindrome
grandyang opened this issue · comments
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.
验证回文字符串是比较常见的问题,所谓回文,就是一个正读和反读都一样的字符串,比如“level”或者“noon”等等就是回文串。但是这里,加入了空格和非字母数字的字符,增加了些难度,但其实原理还是很简单:只需要建立两个指针,left和right, 分别从字符的开头和结尾处开始遍历整个字符串,如果遇到非字母数字的字符就跳过,继续往下找,直到找到下一个字母数字或者结束遍历,如果遇到大写字母,就将其转为小写。等左右指针都找到字母数字时,比较这两个字符,若相等,则继续比较下面两个分别找到的字母数字,若不相等,直接返回false.
时间复杂度为O(n), 代码如下:
解法一:
class Solution {
public:
bool isPalindrome(string s) {
int left = 0, right = s.size() - 1 ;
while (left < right) {
if (!isAlphaNum(s[left])) ++left;
else if (!isAlphaNum(s[right])) --right;
else if ((s[left] + 32 - 'a') %32 != (s[right] + 32 - 'a') % 32) return false;
else {
++left; --right;
}
}
return true;
}
bool isAlphaNum(char &ch) {
if (ch >= 'a' && ch <= 'z') return true;
if (ch >= 'A' && ch <= 'Z') return true;
if (ch >= '0' && ch <= '9') return true;
return false;
}
};
我们也可以用系统自带的判断是否是数母字符的判断函数isalnum,参见代码如下;
解法二:
class Solution {
public:
bool isPalindrome(string s) {
int left = 0, right = s.size() - 1 ;
while (left < right) {
if (!isalnum(s[left])) ++left;
else if (!isalnum(s[right])) --right;
else if ((s[left] + 32 - 'a') %32 != (s[right] + 32 - 'a') % 32) return false;
else {
++left; --right;
}
}
return true;
}
};
对于该问题的扩展,还有利用Manacher算法来求解最长回文字符串问题,参见我的另一篇博文Manacher's Algorithm 马拉车算法。
参考资料:
https://discuss.leetcode.com/topic/5581/here-s-a-clean-c-solution/2
https://discuss.leetcode.com/topic/25405/my-three-line-java-solution