代码随想训练营第五十五天（Python）｜ 392.判断子序列、115.不同的子序列-JZTXT

392.判断子序列
1、双指针

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        m, n = len(s), len(t)
        i, j = 0, 0
        while m > i and n > j:
            if s[i] == t[j]:
                i += 1
            j += 1
        return m == i

2、动态规划

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        m, n = len(s), len(t)
        # dp 数组代表 s 以 i-1 结尾和 t 以 j 结尾的最长公共子序列
        dp = [[0] * (n+1) for _ in range(m+1)]

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

        if dp[m][n] == m:
            return True
        else:
            return False

115.不同的子序列

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        n, m = len(s), len(t)
        # dp 数组代表 s 以 i-1 结尾中包含 t 以 j-1 结尾的个数
        dp = [[0] * (m+1) for _ in range(n+1)]

        # 初始化
        for i in range(n+1):
            dp[i][0] = 1

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

        return dp[n][m]