LeetCode115 Distinct Subsequences

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

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.

分析：

看题目感觉就跟LCS很像，考虑用双序列动态规划解决。

1. 状态：

dp[i][j]表示从第一个字符串前i个组成的子串转换为第二个字符串前j个组成的子串共有多少种方案。

2. 递推：

s[i - 1] == t[j - 1]， 则dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];

s[i - 1] != t[j - 1]，则dp[i][j] = dp[i - 1][j];

3. 初始化：

dp[i][0] = 1（删除到没有字符只有一种方案）

4. 返回值：

dp[sz1 - 1][sz2 - 1]

代码：

class Solution {
public:
    int numDistinct(string s, string t) {
        int sz1 = s.size(), sz2 = t.size();
        int dp[sz1 + 1][sz2 + 1];
        memset(dp, 0 , sizeof(dp));
        for (int i = 0; i < sz1; ++i) {
            dp[i][0] = 1;
        }
        for (int i = 1; i <= sz1; ++i) {
            for (int j = 1; j <= sz2; ++j) {
                if (s[i - 1] == t[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                }
                else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[sz1][sz2];
    }
};