--- 最长公共子序列 (LCS) ---一个序列的子序列是在该序列中删去若干元素后得到的序列例："ABCD"和"BDF"都是"ABCDEFG"的子序列最长公共子序列 (LCS) 问题：给定两个序列 X 和 Y，求 X 和 Y 长度最大的公共子序列例：X = "ABBCBDE", Y = "DBBCDB", LCS(X, Y) = "BBCD"适用场景：1. 字符串相似度比对

def lcs_length(x, y):m, n = len(x), len(y)dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]for i in range(1, m + 1):for j in range(1, n + 1):# i j 位置上的字符匹配时，来自于左上方+1if x[i - 1] == y[j - 1]:dp[i][j] = dp[i - 1][j - 1] + 1else:dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])return dp[m][n]def lcs(x, y):m, n = len(x), len(y)dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]# 记录路径：1. 左上方，2. 上方，3. 左方trace = [[0 for _ in range(n + 1)] for _ in range(m + 1)]for i in range(1, m + 1):for j in range(1, n + 1):# 来自于左上方if x[i - 1] == y[j - 1]:dp[i][j] = dp[i - 1][j - 1] + 1trace[i][j] = 1# 来自于上方elif dp[i - 1][j] > dp[i][j - 1]:dp[i][j] = dp[i - 1][j]trace[i][j] = 2# 来自于左方else:dp[i][j] = dp[i][j - 1]trace[i][j] = 3return dp[m][n], tracedef lcs_traceback(x, y):dp, trace = lcs(x, y)i, j = len(x), len(y)res = []while i and j:# 来自于左上方：匹配if trace[i][j] == 1:res.append(x[i - 1])i -= 1j -= 1# 来自于上方：不匹配elif trace[i][j] == 2:i -= 1# 来自于左方：不匹配else:j -= 1return ''.join(reversed(res))