LeetCode115—Distinct Subsequences

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本文介绍了一种使用动态规划解决字符串子序列计数问题的方法。通过定义状态dp(i,j)来表示母串前i个字符与子串前j个字符的匹配情况，实现了对任意两个字符串间子序列数量的有效计算。

　原题

链接
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.

分析

子序列的个数．母串 $S={s_0s_1s_2 \dots s_{i-2}s_{i-1} \dots}$ ，子串 $T={t_0t_1t_2 \dots t_{i-2}t_{i-1} \dots}$
令 $dp(i,j)$ 表示 $\{s_0s_1s_2 \dots s_{i-2}s_{i-1}\}$ 和 $\{t_0t_1t_2 \dots t_{i-2}t_{i-1}\}$ 的状态
可以列出如下动规方程：

d p (i, j) = {d p (i - 1, j - 1) + d p (i - 1, j) d p (i - 1, j), s (i - 1) = = t (j - 1), s (i - 1) \neq t (j - 1)

$dp(i,j)= \begin{cases} dp(i-1,j-1)+dp(i-1,j) &, s(i-1)==t(j-1)\\ dp(i-1,j)&,s(i-1) \neq t(j-1) \end{cases}$

代码

对照方程写出代码即可

class Solution {
    public:
    int numDistinct(string s, string t)
    {
        int m=s.size();
        int n=t.size();
        vector< vector<int> > dp(m+1,vector<int>(n+1,0));
        for(int i=0;i<=m;++i)
        {
            dp[i][0]=1;
        }
        for(int i = 1;i <= m; ++i)
        {
            for(int j = 1; j <= n; ++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[m][n];
    }
};