125 Valid Palindrome
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
(大意:給定字符串,只考慮其中的數字和字母字符,判斷該字符串是否爲迴文串)
For example:
“A man, a plan, a canal: Panama” is a palindrome.
“race a car” is not a palindrome.
思路:
1. 定義兩個指針,分佈指向字符串的首尾
2. 從頭開始,過濾頭部的非字母數字字符
3. 然後從尾部開始,過尾部的非字母數字字符
4. 頭指針小於尾指針的情況下,
1) 如果兩個字母數字字符相同,
2)頭指針後移,並過濾非字母數字字符
5. 如果兩個字母不同,返回false
代碼:
class Solution {
public:
bool isPalindrome(string s) {
if (s.size() == 0 || s.size() == 1) return true;
int low = 0, high = s.size() - 1;
//從頭開始,過濾頭部的非字母數字字符
while (low < s.size() && !isRightChar(s[low])) ++low;
//從尾開始,過濾尾部的非字母數字字符
while (high >= 0 && !isRightChar(s[high])) --high;
while(low < high){
if (lowerCase(s[low]) == lowerCase(s[high])){
//頭指針後移,並過濾非數字字母字符
++low;
while (low < s.size() && !isRightChar(s[low])) ++low;
//尾指針前移,並過濾非數字字母字符
--high;
while (high >= 0 && !isRightChar(s[high])) --high;
}else return false;
}
return true;
}
//判讀是否爲數字字母字符
bool isRightChar(char c){
if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c >= '0' && c <= '9')) return true;
else return false;
}
//大寫轉小寫
char lowerCase(char c){
if (c >= 'A' && c <= 'Z') return tolower(c);
return c;
}
};