07
05/2015
[LeetCode] 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.
解题思路：
题意为不区分大小写，且只判断数字和字母。有个transform函数学会了：std::transform(s.begin(), s.end(), s.begin(), ::tolower);，能够将字符串转化成小写形式。
class Solution {
public:
bool isPalindrome(string s) {
int len = s.length();
std::transform(s.begin(), s.end(), s.begin(), ::tolower); //这个函数学会
int i=0, j=len-1;
while(i<j){
if(!isAlphanumeric(s[i])){
i++;
continue;
}
if(!isAlphanumeric(s[j])){
j--;
continue;
}
if(s[i]!=s[j]){
return false;
}
i++;
j--;
}
return true;
}
private:
bool isAlphanumeric(char c){
if(c>='a'&&c<='z' || c>='0'&&c<='9'){
return true;
}
return false;
}
};二次刷题（2015-08-05）
class Solution {
public:
bool isPalindrome(string s) {
int i = 0, j = s.length() - 1;
while(i<j){
char left = tolower(s[i]);
while(i<j && !isalpha(left) && !isdigit(left)){
i++;
left = tolower(s[i]);
}
char right = tolower(s[j]);
while(i<j && !isalpha(right) && !isdigit(right)){
j--;
right = tolower(s[j]);
}
if(left!=right){
return false;
}
i++;
j--;
}
return true;
}
};转载请注明：康瑞部落 » [LeetCode] Valid Palindrome
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