41：Valid Palindrome

原创于 2017-02-22 09:04:54 发布 · 235 阅读

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本文介绍了一个C++实现的算法，该算法用于判断一个字符串是否为回文，仅考虑字母和数字字符并忽略大小写。通过双指针技术，从两端向中间遍历比较字符，达到O(n)的时间复杂度和O(1)的空间复杂度。

本题解法代码的思想及编写参考了网址https://github.com/soulmachine/leetcode#leetcode题解题目

题目：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.

解题代码如下：

//时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
        bool isPalindrome(string s) {
        //      transform(s.begin(), s.end(), s.begin(), ::tolower);
                auto left = s.begin(), right = --s.end();
                while (left < right) {
                        if (!::isalnum(*left)) ++left;
                        else if (!::isalnum(*right)) --right;
                        else if (::tolower(*left) != tolower(*right))
                                return false;
                        else { ++left, --right; }
                }
                return true;
        }
};