返回子串长度
def findLongestCommonString(A, B):
m = len(A)
n = len(B)
length = 0
dp = [[0] * (n + 1) for i in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
if A[i - 1] == B[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
if dp[i][j] > length:
length = dp[i][j]
else:
dp[i][j] = 0
return length
返回子串长度,子串
def findLongestCommonString(A, B):
m = len(A)
n = len(B)
length = 0
str = ""
dp = [[0] * (n + 1) for i in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
if A[i - 1] == B[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
if dp[i][j] > length:
length = dp[i][j]
str = A[i - length : i]
else:
dp[i][j] = 0
return length, str
本文深入探讨了寻找两个字符串中最长公共子串的算法实现。通过动态规划的方法,我们详细解释了如何构造二维DP数组来高效地找到最长公共子串的长度及具体子串。适合对字符串匹配算法感兴趣的读者。
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