1 class Solution: 2 def __init__(self): 3 self.dp = [] 4 self.b = [] 5 self.LCStr = '' 6 self.stridx1 = [] 7 self.stridx2 = [] 8 9 def LCS(self,str1,str2): 10 m = len(str1) 11 n = len(str2) 12 self.dp = [[0 for _ in range(n+1)]for _ in range(m+1)] 13 self.b = [[0 for _ in range(n+1)]for _ in range(m+1)] 14 for i in range(m): 15 for j in range(n): 16 if str1[i] == str2[j]: 17 self.dp[i+1][j+1] = self.dp[i][j] + 1 18 self.b[i+1][j+1] = 1 19 else: 20 if self.dp[i+1][j] >= self.dp[i][j+1]: 21 self.dp[i+1][j+1] = self.dp[i+1][j] 22 self.b[i+1][j+1] = 2 23 else: 24 self.dp[i+1][j+1] = self.dp[i][j+1] 25 self.b[i+1][j+1] = 3 26 27 def printLCS(self,str1,str2,i,j): 28 if i == 0 or j == 0: 29 return 30 if self.b[i][j] == 1: 31 self.stridx1.insert(0,i-1) 32 self.stridx2.insert(0,j-1) 33 self.LCStr = str1[i-1] + self.LCStr 34 self.printLCS(str1,str2,i-1,j-1) 35 elif self.b[i][j] == 2: 36 self.printLCS(str1,str2,i,j-1) 37 else: 38 self.printLCS(str1,str2,i-1,j) 39 40 def generateStr(self,str1,str2): 41 k = len(self.LCStr) 42 result = '' 43 result += str1[:self.stridx1[0]] + str2[:self.stridx2[0]] + self.LCStr[0] 44 45 if k >= 2: 46 for i in range(1,k): 47 result += str1[self.stridx1[i-1]+1:self.stridx1[i]] + str2[self.stridx2[i-1]+1:self.stridx2[i]] 48 result += self.LCStr[i] 49 50 result += str1[self.stridx1[k-1]+1:] + str2[self.stridx2[k-1]+1:] 51 return result 52 53 def shortestCommonSupersequence(self, str1: str, str2: str) -> str: 54 self.LCS(str1,str2) 55 self.printLCS(str1,str2,len(str1),len(str2)) 56 if len(self.LCStr) == 0: 57 return str1 + str2 58 else: 59 return self.generateStr(str1,str2)
写得有点复杂了，好在AC了。对我而言，hard难度的题目，能做出来算是不错了，效率上就不能做太高要求了。
本题的主要思想：先求两个字符串的最长公共子序列（LCS），然后再把不是最长公共子序列中的字符按任意顺序拼接在一起。
这里拼接策略是先补充str1中的字符，再补充str2中的字符。