leetcode 730 Count Different Palindromic Subsequences

Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo 10^9 + 7.

A subsequence of a string S is obtained by deleting 0 or more characters from S.

A sequence is palindromic if it is equal to the sequence reversed.

Two sequences A_1, A_2, ... and B_1, B_2, ... are different if there is some i for which A_i != B_i.

Example 1:

Input: 
S = 'bccb'
Output: 6
Explanation: 
The 6 different non-empty palindromic subsequences are 'b', 'c', 'bb', 'cc', 'bcb', 'bccb'.
Note that 'bcb' is counted only once, even though it occurs twice.

 

Example 2:

Input: 
S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba'
Output: 104860361
Explanation: 
There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7.

 

Note:

• The length of S will be in the range [1, 1000].
• Each character S[i] will be in the set {'a', 'b', 'c', 'd'}.

解法1.0

class Solution {
public:
	int countPalindromicSubsequences(string S)
	{
		seq = S;
		string s = "";
		return cntpalinsub(0, s);
	}

	int cntpalinsub(int n, string& s)
	{
		long res = 0;
		string t = s + seq[n];
		if (n == seq.size() - 1)
		{
			if (ispalin(s) && !strs.count(s))
			{
				strs.insert(s);
				res += 1;
			}
			if (ispalin(t) && !strs.count(t))
			{
				strs.insert(t);
				res += 1;
			}
			return res % mod;
		}

		res = cntpalinsub(n + 1, s) + cntpalinsub(n + 1, t);

		return res % mod;
	}

	bool ispalin(string& s)
	{
		if (s.empty()) return false;
		string t = s;
		reverse(t.begin(), t.end());
		return s == t;
	}

private:

	string seq = "";
	set<string> strs;
	int mod = 1000000007;
};

解法2.0

参考大神grandyang大神的解法

class Solution {
public:
    int countPalindromicSubsequences(string S) {
        int n = S.size(), M = 1e9 + 7;
        vector<vector<int>> dp(n, vector<int>(n, 0));
        for (int i = 0; i < n; ++i) dp[i][i] = 1;
        for (int len = 1; len < n; ++len) {
            for (int i = 0; i < n - len; ++i) {
                int j = i + len;
                if (S[i] == S[j]) {
                    int left = i + 1, right = j - 1;
                    while (left <= right && S[left] != S[i]) ++left;
                    while (left <= right && S[right] != S[i]) --right;
                    if (left > right) {
                        dp[i][j] = 2 * dp[i + 1][j - 1] + 2 ;
                    } else if (left == right) {
                        dp[i][j] = 2 * dp[i + 1][j - 1] + 1 ;
                    } else {
                        dp[i][j] = 2 * dp[i + 1][j - 1] - dp[left + 1][right - 1] ;
                    }
                } else {
                    dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1];
                }
                dp[i][j] = (dp[i][j] < 0) ? dp[i][j] + M : dp[i][j] % M;
            }
        }
        return dp[0][n - 1];
    }
};