LeetCode——Distinct Subsequences

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).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

» Solve this problem

先对题意解释一下：

题目给出两个字符串S和T，求S的所有子串中与T相同的有多少个。

子串的定义为：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).

定义f[i][j]表示S[0..i-1]的子串中有f[i][j]个与T[0..j-1]相同。

因此，

当S[i] == T[j]时，f[i+1][j+1] = f[i][j] + f[i][j+1]；

当S[i] != T[j]时，f[i+1][j+1] = f[i][j+1]。

class Solution {
public:
    int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int lS = S.length(), lT = T.length();
        if (lS < lT) {
            return 0;
        }
        int f[lS + 1][lT + 1];
        memset(f, 0, sizeof(int) * (lS + 1) * (lT + 1));
        for (int i = 0; i<= lS; i++) {
            f[i][0] = 1;
        }
        for (int i = 1; i <= lS; i++) {
            for (int j = 1; j <= i && j <= lT; j++) {
                f[i][j] = S[i - 1] == T[j - 1] ? f[i - 1][j - 1] : 0;
                if (i - 1 >= j) {
                    f[i][j] += f[i - 1][j];
                }
            }
        }
        return f[lS][lT];
    }
};