LeetCode剑指 Offer II 097 子序列的数目

LeetCode剑指 Offer II 097 子序列的数目

\(f[i][j]\) 表示 \(s[:i]\) 包含 \(t[:j]\) 子序列的个数

\(s[i] == t[i]\) 时, \(f[i][j] = f[i - 1][j - 1] + f[i - 1][j]\), 当前 \(s[i]\) 用或者不用
\(s[i] != t[i]\) 时, \(f[i][j] = f[i - 1][j]\), 当前 \(s[i]\) 不用

class Solution:
    def numDistinct(self, s: str, t: str) -> int:

        m, n = len(s), len(t)
        f = [[0] * (n + 1) for i in range(m + 1)]
        for i in range(m + 1): f[i][0] = 1

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if s[i - 1] == t[j - 1]:
                    f[i][j] = f[i - 1][j - 1] + f[i - 1][j]
                else:
                    f[i][j] = f[i - 1][j]

        return f[m][n]