Question:
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.
Anwser 1:
class Solution {
public:
bool isStr(char &ch){
if(ch >= '0' && ch <= '9'){
return true;
} else if(ch >= 'a' && ch <= 'z'){
return true;
} else if(ch >= 'A' && ch <= 'Z'){
ch += 32;
return true;
}
return false;
}
bool isPalindrome(string s) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int len = s.length();
if(len == 0){
return true;
}
string str = "";
for(int i = 0; i < len; i++){ // remove illegal char, such as "?" "/" ...
if(isStr(s[i])){
str += s[i];
}
}
len = str.length();
int mid = (len + 1) / 2;
for(int i = 0; i < mid; i++){
if(str[i] != str[len - 1 - i]){ // check front and end char
return false;
}
}
return true;
}
};
Anwser 2:
class Solution {
public:
bool isStr(char &ch){
if(ch >= '0' && ch <= '9'){
return true;
} else if(ch >= 'a' && ch <= 'z'){
return true;
} else if(ch >= 'A' && ch <= 'Z'){
ch += 32;
return true;
}
return false;
}
bool isPalindrome(string s) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int len = s.length();
if(len == 0){
return true;
}
int i = 0;
int j = len - 1;
while(i < j){
if(!isStr(s[i])) {
i++;
} else if(!isStr(s[j])) {
j--;
} else if(s[i++] != s[j--]) {
return false;
}
}
return true;
}
};
注意点:
1) 字母数字判断,注意字母中的字母大小写转换(把大写加32全部转化成小写)
2) 方法2中采用判断删除的方法,效率要高,且不占用另外字符串存储(str)