Hints

• DP

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题面

Difficulty	Time Complexity Limit	Extra-Memory Complexity Limit
Hard	$O(nm)$	$-$

Given two strings $s$ and $t$, return the number of distinct subsequences of $s$ which equals $t$.

The test cases are generated so that the answer fits on a 32-bit signed integer.

Example 1:

Input: s = "rabbbit", t = "rabbit"
Output: 3
Explanation:
As shown below, there are 3 ways you can generate "rabbit" from s.
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.

babgbag babgbag babgbag babgbag babgbag

Constraints:

• $1 \le |s|,\ |t| \le 1000$

• $s$ and $t$ consist of English letters.

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题意

给定两个字符串 $s$ 和 $t$ ，求 $s$ 中有多少个位置不同的子序列与 $t$ 相同。

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题解

为便于描述，设字符串的下标从 $1$ 起，字符串 $s$ 和 $t$ 的长度分别为 $n$ 和 $m$ 。

考虑动态规划，设 $dp[i][j]$ 表示 $s[1,\ i]$ 中 $t[1,\ j]$ 的出现次数，状态转移方程为：

$\begin{cases} dp[i][0] = 1, &(initialize) \\ dp[0][j] = 0, &(initialize) \\ dp[i] = dp[i - 1][j] + dp[i - 1][j - 1], &(s[i] = t[j]) \\ dp[i] = dp[i - 1][j], &(s[i] \ne t[j]) \end{cases}$

则 $dp[n][m]$ 即为答案。时间复杂度为 $O(nm)$ 。

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AC代码

class Solution {
public:
	int numDistinct(const string & s, const string & t) {
		// 初始化
		int n = s.length(), m = t.length();
		int dp[n + 1][m + 1];
		memset(dp, 0, sizeof(dp));
		int mod = 1e9 + 7;
		// 递推
		for (int i = 0; i <= n; i ++) {
			dp[i][0] = 1;
		}
		for (int i = 1; i <= n; i ++) {
			for (int j = 1; j <= i && j <= m; j ++) {
				if (s[i - 1] == t[j - 1]) {
					dp[i][j] = dp[i - 1][j] % mod + dp[i - 1][j - 1] % mod;
				} else {
					dp[i][j] = dp[i - 1][j] % mod;
				}
			}
		}
		// 返回
		return dp[n][m];
	}
};

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