最长公共子序列

前缀型动态规划

def longest_common_seq(s1, s2):
    if not s1 or not s2:
        return 
    m, n = len(s1), len(s2)
    # dp[i][j] = max(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]),当前字符依赖于i-1和j-1,需要补一个状态零
    dp = [[0]*(n+1) for _ in range(m+1)]
    for i in range(1, m+1):
        for j in range(1, n+1):
            if s1[i-1] == s2[j-1]:     # 切记字符串的第i个字符是i-1
                dp[i][j] = dp[i-1][j-1] + 1
            else:
                dp[i][j] = max(dp[i-1][j], dp[i][j-1])
    return dp[m][n]
longest_common_seq("deng", "de")

标签：return,seq,s2,s1,range,公共,序列,最长,dp
来源： https://www.cnblogs.com/demo-deng/p/16589885.html