题目
给定两个字符串A和B,返回两个字符串的最长公共子序列的长度。例如,A=”1A2C3D4B56”,B=”B1D23CA45B6A”,”123456”或者”12C4B6”都是最长公共子序列。
给定两个字符串A和B,同时给定两个串的长度n和m,请返回最长公共子序列的长度。保证两串长度均小于等于300。
测试样例:
“1A2C3D4B56”,10,”B1D23CA45B6A”,12
返回:6
思路
dp[i][j]={dp[i−1][j−1]+1,A[i]=B[j]max{dp[i−1][j],dp[i][j−1]},A[i]≠B[j]
代码
class LCS:
def findLCS(self, A, n, B, m):
# write code here
dp = [[0 for i in range(m)] for _ in range(n)]
for i in range(n):
for j in range(m):
if i == 0 and j == 0:
dp[i][j] = 0 if A[i] != B[j] else 1
elif i == 0:
dp[i][j] = 1 if A[i] == B[j] else dp[i][j - 1]
elif j == 0:
dp[i][j] = 1 if A[i] == B[j] else dp[i - 1][j]
elif A[i] == B[j]:
dp[i][j] = dp[i-1][j-1] + 1
else:
dp[i][j] = max(dp[i-1][j], dp[i][j-1])
return dp[n-1][m-1]