动态规划

DNA序列编辑距离

线性动态规划

$f(i,j)$为dna1前i个字符和dna2的前j个字符相等的最小编辑步骤

$f(i,j) = min(f(i-1,j), f(i,j-1), f(i-1,j-1)) + 1$ 当$s[i-1] != s[j-1]$时

$f(i,j) = f(i-1,j-1)$ 此时 $s[i-1] == s[j-1]$ 不需要进行编辑

def solution(dna1, dna2):
    n = len(dna1)
    m = len(dna2)
    
    f = [[0] * (m + 1) for _ in range(n + 1)]
    
    # 初始化边界条件
    for i in range(n + 1):
        f[i][0] = i
    for j in range(m + 1):
        f[0][j] = j
    
    # 动态规划计算编辑距离
    for i in range(1, n + 1):
        for j in range(1, m + 1):
            if dna1[i - 1] == dna2[j - 1]:
                f[i][j] = f[i - 1][j - 1]
            else:
                f[i][j] = min(f[i - 1][j] + 1, f[i][j - 1] + 1, f[i - 1][j - 1] + 1)
    
    return f[n][m]

if __name__ == "__main__":
    # 你可以添加更多的测试用例
    print(solution("AGCTTAGC", "AGCTAGCT") == 2)
    print(solution("AGCCGAGC", "GCTAGCT") == 4)