leetcode1092

class Solution:
    def __init__(self):
        self.dp = []
        self.b = []
        self.LCStr = ''
        self.stridx1 = []
        self.stridx2 = []

    def LCS(self,str1,str2):
        m = len(str1)
        n = len(str2)
        self.dp = [[0 for _ in range(n+1)]for _ in range(m+1)]
        self.b = [[0 for _ in range(n+1)]for _ in range(m+1)]
        for i in range(m):
            for j in range(n):
                if str1[i] == str2[j]:
                    self.dp[i+1][j+1] = self.dp[i][j] + 1
                    self.b[i+1][j+1] = 1
                else:
                    if self.dp[i+1][j] >= self.dp[i][j+1]:
                        self.dp[i+1][j+1] = self.dp[i+1][j]
                        self.b[i+1][j+1] = 2
                    else:
                        self.dp[i+1][j+1] = self.dp[i][j+1]
                        self.b[i+1][j+1] = 3

    def printLCS(self,str1,str2,i,j):
        if i == 0 or j == 0:
            return
        if self.b[i][j] == 1:
            self.stridx1.insert(0,i-1)
            self.stridx2.insert(0,j-1)
            self.LCStr = str1[i-1] + self.LCStr
            self.printLCS(str1,str2,i-1,j-1)
        elif self.b[i][j] == 2:
            self.printLCS(str1,str2,i,j-1)
        else:
            self.printLCS(str1,str2,i-1,j)

    def generateStr(self,str1,str2):
        k = len(self.LCStr)
        result = ''
        result += str1[:self.stridx1[0]] + str2[:self.stridx2[0]] + self.LCStr[0]

        if k >= 2:
            for i in range(1,k):
                result += str1[self.stridx1[i-1]+1:self.stridx1[i]] + str2[self.stridx2[i-1]+1:self.stridx2[i]]
                result += self.LCStr[i]

        result += str1[self.stridx1[k-1]+1:] + str2[self.stridx2[k-1]+1:]
        return result

    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
        self.LCS(str1,str2)
        self.printLCS(str1,str2,len(str1),len(str2))
        if len(self.LCStr) == 0:
            return str1 + str2
        else:
            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。