LeetCode Valid Palindrome
来源：程序员人生 发布时间：2015-08-07 08:29:20 阅读次数：2218次
1.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.
2.解答
首先是题目的意思是：判断1个句子，正的读和倒着读是不是都1样。比如"abba",就是正读和倒读都1样。这题只是多少疏忽大小写，疏忽非字母和数字。所以"A man, a plan, a canal: Panama"正读和倒读是1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，1个指向前，1个指向后，进行比较，过滤掉非字母数字。
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