Given a string S and a string T, count the number of distinct subsequences of S which equals T.
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 1:
Input: S ="rabbbit", T ="rabbit"Explanation: As shown below, there are 3 ways you can generate "rabbit" from S.
Output: 3
(The caret symbol ^ means the chosen letters)rabbbit
^^^^ ^^rabbbit
^^ ^^^^rabbbit
^^^ ^^^
Example 2:
Input: S ="babgbag", T ="bag"Explanation: As shown below, there are 5 ways you can generate "bag" from S.
Output: 5
(The caret symbol ^ means the chosen letters)babgbag
^^ ^babgbag
^^ ^babgbag
^ ^^babgbag
^ ^^babgbag
^^^
看到有关字符串的子序列或者配准类的问题，首先应该考虑的就是用动态规划 Dynamic Programming 来求解，这个应成为条件反射。而所有 DP 问题的核心就是找出状态转移方程，想这道题就是递推一个二维的 dp 数组，其中 dp[i][j] 表示s中范围是 [0, i] 的子串中能组成t中范围是 [0, j] 的子串的子序列的个数。下面我们从题目中给的例子来分析，这个二维 dp 数组应为：
Ø r a b b b i t
Ø 1 1 1 1 1 1 1 1
r 1 1
a 1 1
b 0 1 2
b 1 3
i 0 3
t 0 3
首先，若原字符串和子序列都为空时，返回1，因为空串也是空串的一个子序列。若原字符串不为空，而子序列为空，也返回1，因为空串也是任意字符串的一个子序列。而当原字符串为空，子序列不为空时，返回0，因为非空字符串不能当空字符串的子序列。理清这些，二维数组 dp 的边缘便可以初始化了，下面只要找出状态转移方程，就可以更新整个 dp 数组了。我们通过观察上面的二维数组可以发现，当更新到 dp[i][j] 时，dp[i][j] >= dp[i][j - 1] 总是成立，再进一步观察发现，当 T[i - 1] == S[j - 1] 时，dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]，若不等， dp[i][j] = dp[i][j - 1]，所以，综合以上，递推式为：
dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)
根据以上分析，可以写出代码如下：
class Solution {
public:
int numDistinct(string s, string t) {
int m = s.size(), n = t.size();
vector<vector<long>> dp(n + , vector<long>(m + ));
for (int j = ; j <= m; ++j) dp[][j] = ;
for (int i = ; i <= n; ++i) {
for (int j = ; j <= m; ++j) {
dp[i][j] = dp[i][j - ] + (t[i - ] == s[j - ] ? dp[i - ][j - ] : );
}
}
return dp[n][m];
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/115
参考资料：
https://leetcode.com/problems/distinct-subsequences/
https://leetcode.com/problems/distinct-subsequences/discuss/37327/Easy-to-understand-DP-in-Java