leetcode | Valid Palindrome

来源：互联网发布：嗨森矩阵编辑：程序博客网时间：2024/05/12 10:03

Valid Palindrome : https://leetcode.com/problems/valid-palindrome/

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.

解析

alphanumeric characters ：字母或数字型字符

本题难点在于：
1. 全空格字符串应判断为true
2. 字母不区分大小写

一种算法就是，将原字符串剔除所有非字母数字字符，插入到一个新的字符串空间，然后基于新字符串，用2个指针法做判断。算法实现简单，不需额外考虑全空格字符串，但是占用额外空间。

另一种算法是，用2个指针指向字符串两头，当遇到非字母数字字符时跳过，然后做处理。对于全空格字符串，通过判断左指针是否会到达字符串尾部。，优点是不占用额外空间，时间复杂度上也优于第一种（只需遍历一次数组）。

算法 1 实现

class Solution {public:    bool isPalindrome(string s) {        string s2;        int i = -1;        while (++i < s.size()) {            if ((s[i] < 'A' || s[i] > 'Z') && (s[i] < 'a' || s[i] > 'z') && (s[i] < '0' || s[i] > '9'))                continue;            else                s2 += s[i];        }        i = -1;        int j = s2.size();        if (j == 0)            return true;        int interval = 'a' - 'A';        while (++i <= --j) {            if (!(s2[i] == s2[j] || s2[i]+interval == s2[j] || s2[i]-interval == s2[j]))                return false;        }        return true;    }};

算法 2 实现

class Solution {public:    bool isPalindrome(string s) {        if (s.size() == 0)            return true;        int i = -1;        int j = s.size();        int interval = 'a' - 'A';        while (++i <= --j) {            while ((i < j) && (s[i] < 'A' || s[i] > 'Z') && (s[i] < 'a' || s[i] > 'z') && (s[i] < '0' || s[i] > '9'))                i++;            while ((i < j) && (s[j] < 'A' || s[j] > 'Z') && (s[j] < 'a' || s[j] > 'z') && (s[j] < '0' || s[j] > '9'))                j--;            // 全空格字符串            if (i == s.size()-1)                return true;            if (s[i] != s[j] && s[i]+interval != s[j] && s[i]-interval != s[j])                return false;        }        return true;    }};

0 0