计算字符串距离

解题思路:

注意事项:

参考代码:

def edit_distance(A, B):

    m = len(A)

    n = len(B)

    # 初始化二维数组

    dp = [[0] * (n + 1) for _ in range(m + 1)]

    # 边界情况处理

    for i in range(m + 1):

        dp[i][0] = i

    for j in range(n + 1):

        dp[0][j] = j

    # 动态规划计算最小操作次数

    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]

            else:

                dp[i][j] = min(dp[i - 1][j - 1] + 1, dp[i - 1][j] + 1, dp[i][j - 1] + 1)

    return dp[m][n]

# 读取输入

n = int(input())

test_cases = []

for _ in range(n):

    A, B = input().split()

    test_cases.append((A, B))

# 计算并输出结果

for A, B in test_cases:

    result = edit_distance(A, B)

    print(result)