代码随想录算法训练营Day55 (Day 54休息) | 动态规划(15/17) LeetCode 392.判断子序列 115.不同的子序列

继续子序列的练习！

第一题

392. Is Subsequence

Given two strings s and t, return trueif sis a subsequence of t, or falseotherwise.

A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "++a++ b++c++ d++e++" while "aec" is not).

首先想到双指针的解法，复杂度为O(n)，也能接受。不过既然在练习动态规划，就还是按照动态规划的思路去解。

在确定递推公式的时候，首先要考虑如下两种操作，整理如下：

• if (si - 1 == tj - 1)
  • t中找到了一个字符在s中也出现了
• if (si - 1 != tj - 1)
  • 相当于t要删除元素，继续匹配

if (si - 1 == tj - 1)，那么dpij = dpi - 1j - 1 + 1;，因为找到了一个相同的字符，相同子序列长度自然要在dpi-1j-1的基础上加1

if (si - 1 != tj - 1)，此时相当于t要删除元素，t如果把当前元素tj - 1删除，那么dpij 的数值就是 看si - 1与 tj - 2的比较结果了，即：dpij = dpij - 1;

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        dp = [[0] * (len(t)+1) for _ in range(len(s)+1)]
        for i in range(1, len(s)+1):
            for j in range(1, len(t)+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[-1][-1] == len(s):
            return True
        return False

第二题

115. Distinct Subsequences

Given two strings s and t, return the number of distinct subsequences of swhich equalst.

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

这道题双指针就没法做了，只能用动态规划。

递推公式为：dpij = dpi - 1j;

从递推公式dpij = dpi - 1j - 1 + dpi - 1j; 和 dpij = dpi - 1j; 中可以看出dpij 是从上方和左上方推导而来，那么 dpi0 和dp0j是一定要初始化的。

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        n1, n2 = len(s), len(t)
        if n1 < n2:
            return 0

        dp = [0 for _ in range(n2 + 1)]
        dp[0] = 1

        for i in range(1, n1 + 1):

            prev = dp.copy()
            end = i if i < n2 else n2
            for j in range(1, end + 1):
                if s[i - 1] == t[j - 1]:
                    dp[j] = prev[j - 1] + prev[j]
                else:
                    dp[j] = prev[j]
        return dp[-1]