题目来源
Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.
For example:
Given "aacecaaa", return "aaacecaaa".
Given "abcd", return "dcbabcd".
第一想法就是找最长的0-i的回文，然后补上后面的就行。
然后进一步的想法就是找字符串中第一个字母出现的位置，判断其是不是回文。
如下，但是超时了。
class Solution {
public:
string shortestPalindrome(string s) {
int n = s.length();
if (n == 0)
return "";
int cut = 0;
for (int i=1; i<n; i++) {
if (s[i] == s[0] && isPalindrome(s.substr(0, i+1))) {
cut = max(cut, i);
}
}
string t = s.substr(cut+1);
reverse(t.begin(), t.end());
return t + s;
}
bool isPalindrome(string s)
{
int l = 0, r = s.length()-1;
while (l < r) {
if (s[l] == s[r]) {
l++;
r--;
}
else
return false;
}
return true;
}
};
看了下答案，巧妙的利用了KMP算法算出最长的0-i的回文，只需要O(n)的复杂度。什么是KMP？可以看看阮一峰老师的博客。
class Solution {
public:
string shortestPalindrome(string s) {
int n = s.length();
if (n == 0)
return "";
string rev_s = s;
reverse(rev_s.begin(), rev_s.end());
string new_s = s + '#' + rev_s;
vector<int> p(2*n+1, 0);
for (int i=1; i<2*n+1; i++) {
int j = p[i-1];
while (j > 0 && new_s[i] != new_s[j])
j = p[j-1];
p[i] =(j += new_s[i] == new_s[j]);
}
string t = rev_s.substr(0, n-p[2*n]);
return t + s;
}
};
