|-时间复杂度O(m*n)
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python源代码:
class Solution(object):
# 1 回溯法
def back(self, S1, S2):
m = len(S1)-1
n = len(S2)-1
# 终止条件
if m < 0 or n < 0:
return 0
# 递归过程
if S1[m] == S2[n]:
return 1+self.back(S1[:m], S2[:n])
else:
# 不相等
return max(self.back(S1[:m], S2), self.back(S1, S2[:n]))
# 2 结构化子函数
def childS(self, S1, S2):
m = len(S1) - 1
n = len(S2) - 1
rows = [-1 for i in range(n + 1)]
memo = [rows.copy() for j in range(m + 1)]
return self.struct(S1, S2, memo)
def struct(self, S1, S2, memo):
m = len(S1) - 1
n = len(S2) - 1
# 终止条件
if m < 0 or n < 0:
return 0
if S1[m] == S2[n]:
if memo[m-1][n-1] == -1:
memo[m - 1][n - 1] = self.struct(S1[:m], S2[:n], memo)+1
return memo[m-1][n-1]
else:
if memo[m-1][n] == -1:
memo[m - 1][n] = self.struct(S1[:m], S2, memo)
if memo[m][n-1] == -1:
memo[m][n - 1] = self.struct(S1, S2[:n], memo)
return max(memo[m][n - 1], memo[m-1][n])
# 3 动态规划
def dp(self, S1, S2):
m = len(S1)
n = len(S2)
if m < 0 or n < 0:
return 0
memo = [[0]*(n+1) for j in range(m+1)]
# 初始状态 第0行 第0列 都是0
for i in range(1, m+1):
for j in range(1, n+1):
if S1[i-1] == S2[j-1]: # S1中的第i个字符 S2中的第j个字符
memo[i][j] = 1 + memo[i-1][j-1]
else:
memo[i][j] = max(memo[i-1][j], memo[i][j-1])
return memo[m][n]
S1 = 'AFDSAFDSA'
S2 = 'FDSAFD'
print(Solution().dp(S1, S2))class Solutio