leetcode:交错字符串(python)

乖乖的函数 2020-11-13 02:53:28
LeetCode 字符 字符串 错字 交错


1. 题目描述

给定三个字符串 s1, s2, s3, 验证 s3 是否是由 s1 和 s2 交错组成的。
示例 1:

输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出: true

示例 2:

输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出: false

2. 思路

2.1 递归

先检查s3的第一个字符,如果等于s1的第一个字符,且不等于s2的第二个字符,将s1和s3向后移动一位,递归。当等于s2的第一个字符且不等于与s1的第一个字符的时候,将s2和s3向后移动一位,递归。如果即等于s1的第一个字符又等于s2的第一个字符,就有可能有两种情况,s2不动,s1向后移动一位或者是s1不动,s2向后移动一位。

不过这种方法最会有一个很长的字符串,让你运行超时

代码

class Solution:
def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
if len(s3) != len(s1) + len(s2):
return False
if s1 == "" and s2 == "" and s3 == "":
return True
if s1 == "":
return s2 == s3
if s2 == "":
return s1 == s3
if s3[0] == s1[0] and s3[0] != s2[0]:
return self.isInterleave(s1[1:],s2,s3[1:])
elif s3[0] == s2[0] and s3[0] != s1[0]:
return self.isInterleave(s1,s2[1:],s3[1:])
elif s3[0] == s2[0] and s3[0] == s1[0]:
return self.isInterleave(s1[1:],s2,s3[1:]) or self.isInterleave(s1,s2[1:],s3[1:])
else:
return False

2.2 动态规划

dp[i][j] 表示用 s1的前 (i+1)和 s2 的前 (j+1) 个字符,总共 (i+j+2)个字符,是否交错构成 s3的前缀。为了求出 dp[i][j],我们需要考虑 2 种情况:

  1. 当s1 的第 i个字符和 s2的第 j 个字符都不能匹配 s3的第 k个字符,其中 k=i+j 。这种情况下,s1 和 s2 的前缀无法交错形成 s3 长度为 k 的前缀。因此,我们让 dp[i][j] 为 False。
  2. s1 的第 i个字符或者 s2 的第 j 个字符可以匹配 s3的第 kk 个字符,其中 k=i+j 。假设匹配的字符是 x且与 s1的第 i 个字符匹配,我们就需要把 x放在已经形成的交错字符串的最后一个位置。此时,为了我们必须确保 s1 的前 (i-1) 个字符和 s2 的前 j个字符能形成 s3的一个前缀。类似的,如果我们将 s2的第 j个字符与 s3的第 k 个字符匹配,我们需要确保 s1的前 i个字符和 s2的前 (j-1)个字符能形成 s3 的一个前缀,我们就让 dp[i][j] 为 True。
要注意第一行和第一列的初始化

代码

class Solution:
def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
if len(s3) != len(s1) + len(s2):
return False
dp = [[True for j in range(len(s1)+1)] for i in range(len(s2)+1)]
for i in range(len(s2)+1):
for j in range(len(s1)+1):
if i == 0 and j == 0:
dp[i][j] = True
elif i == 0:
dp[i][j] = dp[i][j-1] and s1[j-1] == s3[j-1]
elif j == 0:
dp[i][j] = dp[i-1][j] and s2[i-1] == s3[i-1]
else:
dp[i][j] = (dp[i-1][j] and s2[i-1] == s3[i+j-1]) or (dp[i][j-1] and s1[j-1] == s3[i+j-1])
return dp[-1][-1]

优化

和其他的动态规划一样,仍然可以用一个一位数组来优化二维数组。

优化代码

class Solution:
def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
if len(s3) != len(s1) + len(s2):
return False
dp = [True for i in range(len(s1)+1)]
for i in range(1,len(s1)+1):
dp[i] = dp[i-1] and s1[i-1] == s3[i-1]
for i in range(len(s2)):
for j in range(len(s1)+1):
if j == 0:
dp[j] = dp[j] and s2[i] == s3[i]
else:
dp[j] = (dp[j-1] and s1[j-1] == s3[i+j]) or (dp[j] and s2[i] == s3[i+j])
return dp[-1]
版权声明
本文为[乖乖的函数]所创,转载请带上原文链接,感谢
https://blog.csdn.net/ggdhs/article/details/93868021

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