[LeetCode] 115. 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
    ^^^

 

看到有關字符串的子序列或者配准類的問題，首先應該考慮的就是用動態規划 Dynamic Programming 來求解，這個應成為條件反射。而所有 DP 問題的核心就是找出狀態轉移方程，想這道題就是遞推一個二維的 dp 數組，其中 dp[i][j] 表示s中范圍是 [0, i] 的子串中能組成t中范圍是 [0, j] 的子串的子序列的個數。下面我們從題目中給的例子來分析，這個二維 dp 數組應為：

 

  Ø r a b b b i t
Ø 1 1 1 1 1 1 1 1
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為空時，返回1，因為空串也是空串的一個子序列。若原字符串不為空，而子序列為空，也返回1，因為空串也是任意字符串的一個子序列。而當原字符串為空，子序列不為空時，返回0，因為非空字符串不能當空字符串的子序列。理清這些，二維數組 dp 的邊緣便可以初始化了，下面只要找出狀態轉移方程，就可以更新整個 dp 數組了。我們通過觀察上面的二維數組可以發現，當更新到 dp[i][j] 時，dp[i][j] >= dp[i][j - 1] 總是成立，再進一步觀察發現，當 T[i - 1] == S[j - 1] 時，dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]，若不等， dp[i][j] = dp[i][j - 1]，所以，綜合以上，遞推式為：

dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)

根據以上分析，可以寫出代碼如下：

 

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

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/115

 

參考資料：

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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