题目链接
tag:
- Hard；
question：
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"
Output: 3
Explanation:
As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)
rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^
Example 2:
Input: S = "babgbag", T = "bag"
Output: 5
Explanation:
As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)
babgbag
^^ ^
babgbag
^^ ^
babgbag
^ ^^
babgbag
^ ^^
babgbag
^^^
思路：
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递推公式，想这道题就是递推一个二维的dp数组，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 11
r 0 1 1 1 1 1 1 1
a 0 0 1 1 1 1 1 1
b 0 0 0 1 2 3 3 3
b 0 0 0 0 1 3 3 3
i 0 0 0 0 0 0 3 3
t 0 0 0 0 0 0 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，下面只要找出递推式，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)
代码如下：
class Solution {
public:
int numDistinct(string s, string t) {
if (s.size() < t.size())
return 0;
int n1=t.size(), n2=s.size();
long dp[n1+1][n2+1];
for (int i=0; i<=n1; ++i)
dp[i][0] = 0;
for (int i=0; i<=n2; ++i)
dp[0][i] = 1;
for (int i=1; i<=n1; ++i) {
for (int j=1; j<=n2; ++j) {
if (t[i-1] == s[j-1])
dp[i][j] = dp[i][j-1] + dp[i-1][j-1];
else
dp[i][j] = dp[i][j-1];
}
}
return dp[n1][n2];
}
};
