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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.
首先是题目的意思是:判断一个句子,正的读和倒着读是否都一样。比如"abba",就是正读和倒读都一样。这题只是多少忽略大小写,忽略非字母和数字。所以"A man, a plan, a canal: Panama"正读和倒读是一样的。
class Solution {
public:
bool isValideChar(char c){
if(c >= ‘A‘ && c <= ‘Z‘){
return true;
}
if(c >= ‘a‘ && c <= ‘z‘){
return true;
}
if(c >= ‘0‘ && c <= ‘9‘){
return true;
}
return false;
}
char tolowerCase(char c){
if(c >= ‘A‘ && c <= ‘Z‘){
return c - ‘A‘ + ‘a‘;
}else{
return c;
}
}
bool isPalindrome(string s) {
int left = 0;
int right = s.size() - 1;
if(s.size() == 0){
return true;
}
int sSize = s.size();
while((isValideChar(s[left]) == false) && left < s.size()){
++left;
}
while((isValideChar(s[right]) == false) && right >= 0){
--right;
}
while(left < right){
if(tolowerCase(s[left]) == tolowerCase(s[right])){
++left;
while((isValideChar(s[left]) == false) && left < s.size()){
++left;
}
--right;
while((isValideChar(s[right]) == false) && right >= 0){
--right;
}
}else{
return false;
}
}
return true;
}
};还是比较容易的,一个指向前,一个指向后,进行比较,过滤掉非字母数字。
http://www.waitingfy.com/archives/1680
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原文地址:http://blog.csdn.net/fox64194167/article/details/45420559