Leetcode Palindrome Partitioning

Palindrome Partitioning

Total Accepted:4585Total Submissions:18745My Submissions

Given a strings, partitionssuch that every substring of the partition is a palindrome.

Return all possible palindrome partitioning ofs.

For example, givens="aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

典型的递归回溯法，当然可以利用动态规划法提高点效率。

下面是标准的递归回溯程序：

vector<vector<string> > partition(string s) 
	{
		vector<vector<string> > rs;
		vector<string> tmp;
		storePatition(rs, tmp, s);
		return rs;
	}

	void storePatition(vector<vector<string> > &rs, vector<string> &tmp,
		string &s)
	{
		if (s.empty())
		{
			rs.push_back(tmp);
			return;
		}
		for (int i = 1; i <= s.length(); i++)
		{
			string a = s.substr(0, i);
			if (isPalindrome(a))
			{
				tmp.push_back(a);
				string b = s.substr(i);
				storePatition(rs, tmp, b);
				tmp.pop_back();
			}
		}
	}

	bool isPalindrome(string &s)
	{
		for (int i = 0, j = s.length()-1; i < j; i++, j--)
		{
			if (s[i] != s[j]) return false;
		}
		return true;
	}

//2014-2-18 update
	vector<vector<string> > partition(string s) 
	{
		vector<vector<bool> > table(s.length(), vector<bool>(s.length(), true));
		genTable(table, s);
		vector<vector<string> > rs;
		vector<string> tmp;
		part(rs, tmp, table, s);
		return rs;
	}
	void part(vector<vector<string> > &rs, vector<string> &tmp, 
		vector<vector<bool> > &table, string &s, int start=0)
	{
		if (start == s.length())
		{
			rs.push_back(tmp);
			return;
		}
		for (int d = 1; d <= s.length()-start; d++)
		{
			if (table[start][start+d-1])
			{
				tmp.push_back(s.substr(start, d));
				part(rs, tmp, table, s, start+d);
				tmp.pop_back();
			}
		}
	}
	void genTable(vector<vector<bool> > &table, string &s)
	{
		for (int d = 2; d <= s.length(); d++)
		{
			for (int i = 0, j = d - 1; j < s.length(); i++, j++)
			{
				table[i][j] = (table[i+1][j-1] && s[i] == s[j]);
			}
		}
	}