[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(guān)字符串的子序列或者配準(zhǔn)類的問題，首先應(yīng)該考慮的就是用動(dòng)態(tài)規(guī)劃 Dynamic Programming 來求解，這個(gè)應(yīng)成為條件反射。而所有 DP 問題的核心就是找出狀態(tài)轉(zhuǎn)移方程，想這道題就是遞推一個(gè)二維的 dp 數(shù)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(gè)數(shù)。下面我們從題目中給的例子來分析，這個(gè)二維 dp 數(shù)組應(yīng)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(shí)，返回1，因?yàn)榭沾彩强沾囊粋€(gè)子序列。若原字符串不為空，而子序列為空，也返回1，因?yàn)榭沾彩侨我庾址囊粋€(gè)子序列。而當(dāng)原字符串為空，子序列不為空時(shí)，返回0，因?yàn)榉强兆址荒墚?dāng)空字符串的子序列。理清這些，二維數(shù)組 dp 的邊緣便可以初始化了，下面只要找出狀態(tài)轉(zhuǎn)移方程，就可以更新整個(gè) dp 數(shù)組了。我們通過觀察上面的二維數(shù)組可以發(fā)現(xiàn)，當(dāng)更新到 dp[i][j] 時(shí)，dp[i][j] >= dp[i][j - 1] 總是成立，再進(jìn)一步觀察發(fā)現(xiàn)，當(dāng) T[i - 1] == S[j - 1] 時(shí)，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(jù)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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