题目:
Given a non-empty string s
, you may delete at most one
character. Judge whether you can make it a palindrome.
Example 1:
Input: "aba" Output: True
Example 2:
Input: "abca" Output: True Explanation: You could delete the character 'c'.
Note:
- The string will only contain lowercase characters a-z. The maximum length of the string is 50000.
思路:
我们采用递归的思路:如果s[left] = s[right],那么递归检查内部就可以;否则我们必须删除掉s[left]或者s[right],如果删除其中一个之后形成了回文串就说明可以;否则说明无法通过删除掉其中一个来生成回文串。
代码:
class Solution {
public:
bool validPalindrome(string s) {
int length = s.length();
if (length == 0) {
return true;
}
else {
return validPalindrome(s, 0, length - 1);
}
}
private:
bool validPalindrome(string &s, int left, int right) {
if (left >= right) {
return true;
}
if (s[left] == s[right]) { // we do not need to remove
return validPalindrome(s, left + 1, right - 1);
}
else { // we have to remove s[left] or s[right]
return isPalindrome(s, left + 1, right) || isPalindrome(s, left, right - 1);
}
}
bool isPalindrome(string &s, int left, int right) {
while (left < right) {
if (s[left++] != s[right--]) {
return false;
}
}
return true;
}
};