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C++实现LeetCode(115.不同的子序列)

运维开发网 https://www.qedev.com 2021-07-26 15:31 出处:网络 作者: Grandyang
[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.

[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 编程客栈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 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编程客栈 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,PtNfx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 + 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) {
            PtNfx    dp[i][j] = dp[i][j - 1] + (t[i - 1] == s[j - 1] ? dp[i - 1][j PtNfx- 1] : 0);
            }
        }
        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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