[Leetcode] 125. Valid Palindrome 解题报告

最新推荐文章于 2024-10-19 19:09:29 发布

魔豆Magicbean

最新推荐文章于 2024-10-19 19:09:29 发布

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分类专栏： IT公司面试习题文章标签： Leetcode 解题报告 Two pointers

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本文介绍了一种使用C++实现的高效方法来判断一个字符串是否为回文，特别考虑了只包含字母和数字的情况，并忽略大小写。通过双指针技巧遍历字符串两端，跳过非字母数字字符进行比较。

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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.

思路：

这是一道名副其实的水题。然而如果了解了STL中的库函数之后，会让你写起来更得心应手。这里总结几个STL中的相关函数吧：

int isdigit(int c): check if a character is decimal digit. Decimal digits are any of: 0 1 2 3 4 5 6 7 8 9

int isalpha(int c): check if a character is alphabetic. Checks whether c is an alphabetic letter, either uppercase or lowercase.

int isalnum(int c): check if a character is alphanumeric. Checks whether c is either a decimal digit or an uppercase or lowercase letter.

代码：

class Solution {
public:
    bool isPalindrome(string s) {
        int length = s.length();
        int left = 0, right = s.size() - 1;
        while (left < right) {
            while (left < right && !isalnum(s[left])) {
                ++left;
            }
            while (left < right && !isalnum(s[right])) {
                --right;
            }
            if (left < right && tolower(s[left++]) != tolower(s[right--])) {
                return false;
            }
        }
        return true;
    }
};