【Lintcode】118.Distinct Subsequences

题目：

Given a string S and a string T, count the number of distinct subsequences of T in S.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

Example

Given S = "rabbbit", T = "rabbit", return 3.

题解：

Solution 1 ()

class Solution {
public:
    int numDistinct(string &S, string &T) {
        int n1 = S.size(), n2 = T.size();
        vector<vector<int>> dp(n2 + , vector<int>(n1 + , ));
        for (int i = ; i <= n1; ++i) {
            dp[][i] = ;
        }
        for (int i = ; i <= n2; ++i) {
            for (int j = ; j <= n1; ++j) {
                if (T[i - ] == S[j - ]) {
                    dp[i][j] = dp[i][j - ] + dp[i - ][j - ];
                } else {
                    dp[i][j] = dp[i][j - ];
                }
            }
        }
        return dp[n2][n1];
    }
};

　　Notice that we keep the whole m*n matrix simply for dp[i - 1][j - 1]. So we can simply store that value in a single variable and further optimize the space complexity. The final code is as follows.

Solution 2 ()  from here

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = t.length(), n = s.length();
        vector<int> cur(m + , );
        cur[] = ;
        for (int j = ; j <= n; j++) {
            int pre = ;
            for (int i = ; i <= m; i++) {
                int temp = cur[i];
                cur[i] = cur[i] + (t[i - ] == s[j - ] ? pre : );
                pre = temp;
            }
        }
        return cur[m];
    }
};