leetcode 214. Shortest Palindrome

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".

class Solution {
public:
	string shortestPalindrome(string s) {
		string str;
		if (s.empty() || s.length() == 1)
			return str;
		if (s.length() == 2)
		{
			if (s[0] != s[1])
				s.insert(s.begin(), s[1]);
			return s;
		}
		if (s.length() == 3)
		{
			if (s[0] != s[2])
			{
				s.insert(s.begin(), s[1]);
				s.insert(s.begin(), s[3]);
			}
			return s;
		}
		int min = 10000;
		bool f = false;
		for (int i = s.length() / 2; i > 0; i--)
		{
			int k = 0;
			while (i - k - 1 >= 0 && s[i - k - 1] == s[i + k])
				k++;
			if (i - k == 0 && s[i - 1] == s[i])
			{
				f = true;
				string ss = s;
				string sss(s.begin() + i + k, s.end());
				if (sss.empty())
					return s;
				for (int j = 0; j <= sss.size() / 2; j++)
				{
					char aa = sss[j];
					sss[j] = sss[sss.size() - 1 - j];
					sss[sss.size() - 1 - j] = aa;
				}
				ss.insert(0, sss.c_str());
				if (ss.length() < min)
				{
					min = ss.length();
					str = ss;
				}
			}
			k = 1;
			while (i - k >= 0 && i + k < s.length() && s[i - k] == s[i + k])
				k++;
			if (i - k == -1 && s[i - 1] == s[i + 1])
			{
				f = true;
				string ss = s;
				string sss(s.begin() + i + k, s.end());
				if (sss.empty())
					return s;
				for (int j = 0; j <= sss.size() / 2; j++)
				{
					char aa = sss[j];
					sss[j] = sss[sss.size() - 1 - j];
					sss[sss.size() - 1 - j] = aa;
				}
				ss.insert(0, sss.c_str());
				if (ss.length() < min)
				{
					min = ss.length();
					str = ss;
				}
			}
			if (f)
				return str;
		}
		string ss = s;
		string sss(s.begin() + 1, s.end());
		for (int j = 0; j <= sss.size() / 2; j++)
		{
			char aa = sss[j];
			sss[j] = sss[sss.size() - 1 - j];
			sss[sss.size() - 1 - j] = aa;
		}
		ss.insert(0, sss.c_str());
		return ss;
	}
};