Geeks面试题： Longest Palindromic Subsequence

Given a sequence, find the length of the longest palindromic subsequence in it. For example, if the given sequence is “BBABCBCAB”, then the output should be 7 as “BABCBAB” is the longest palindromic subseuqnce in it.“BBBBB” and “BBCBB” are also palindromic subsequences of the given sequence, but not the longest ones.

The naive solution for this problem is to generate all subsequences of the given sequence and find the longest palindromic subsequence. This solution is exponential in term of time complexity. Let us see how this problem possesses both important properties of a Dynamic Programming (DP) Problem and can efficiently solved using Dynamic Programming.

注意：

这里是最大子序列的palindromic，不是最大子串的palindromi（leetCode上有道题是）。

最大子序列就是：

1 所选取的字符是原数列的字符

2 字符可以间隔抽去（主要区别，子串是不能有间隔的）

3 字符的顺序不能打乱。

int lpsStrDP(string &s)
{
	int n = s.length();
	vector<vector<int> > ta(n, vector<int>(n));
	for (int i = 0; i < n; i++)
	{
		ta[i][i] = 1;
	}

	for (int d = 2; d <= n; d++)
	{
		for (int i = 0; i < n-d+1; i++)
		{
			int j = i+d-1;
			if (d==2 && s[i] == s[j])
				ta[i][j] = 2;
			else if (s[i] == s[j])
				ta[i][j] = ta[i+1][j-1] +2;
			else ta[i][j] = max(ta[i+1][j], ta[i][j-1]);
		}
	}
	return ta[0][n-1];
}

// Returns the length of the longest palindromic subsequence in seq
int lps(char *seq, int i, int j)
{
	// Base Case 1: If there is only 1 character
	if (i == j)
		return 1;

	// Base Case 2: If there are only 2 characters and both are same
	if (seq[i] == seq[j] && i + 1 == j)
		return 2;

	// If the first and last characters match
	if (seq[i] == seq[j])
		return lps (seq, i+1, j-1) + 2;

	// If the first and last characters do not match
	return max( lps(seq, i, j-1), lps(seq, i+1, j) );
}

int main()
{
	char seq[] = "EEKSFORGEEK";
	int n = strlen(seq);
	printf ("The lnegth of the LPS is %d", lps(seq, 0, n));

	string str = seq;
	cout<<lpsStrDP(str)<<endl;

	system("pause");
	return 0;
}