[LeetCode] Distinct Subsequences 不同的子序列

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
^^^

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

https://leetcode.com/problems/distinct-subsequences/discuss/37412/Any-better-solution-that-takes-less-than-O(n2)-space-while-in-O(n2)-time

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