Description:
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
Note:
For the purpose of this problem, we define empty string as valid palindrome
solution 1
思路很簡單,利用兩個下標,一個從前一個從後,依次往前比較,但是超時了。
class Solution {
public:
if ((s[j] >= 65 && s[j] <= 90) || (s[j] >= 97 && s[j] <= 122))
{
if (s[i] == s[j] || s[i] - s[j] == 32 || s[j] - s[i] == 32)
{
i++; j--;
continue;
}
else
return 0;
}
else if (s[j] == ' ' || !isalnum(s[j]))
j--;
}
else if (s[i] == ' '||!isalnum(s[i]))
i++;
else
return 0;
}
return 1;
}
};
solution 2
思路不變,用一個tolower來轉換大小寫,isalnum()判斷是否爲字母
class Solution {
public:
bool isPalindrome(string s) {
int j = s.length() - 1;
if (j == -1 || j == 0)
return 1;
else
{
int i = 0;
while (i < j)
{
char a=tolower(s[i]);
char b=tolower(s[j]);
if(a==b)
{
i++;j--;
continue;
}
else if(a==' '||!isalnum(a))
i++;
else if(b==' '||!isalnum(b))
j--;
else
return 0;
}
return 1;
}
}
};
做出來之後有一種蜜汁鬱悶。
被整自閉了。
用例“lbl”,我記住了!