题解 | #计算字符串的编辑距离#

# 20240919
while True:
    try:
        s1 = input()
        s2 = input()
        m = len(s1)
        n = len(s2)
        dp = [[1 for j in range(n+1)] for i in range(m+1)] # dp初始化的值要设置为1 
        for j in range(n+1):    # 处理第0行
            dp[0][j] = j
        for i in range(m+1):    # 处理第0列
            dp[i][0] = i
        for i in range(1,m+1):
            for j in range(1,n+1):
                if s1[i-1] == s2[j-1]:  # 注意这里为什么比较的是i-1和j-1的值
                    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)
        print(dp[m][n])
        
    except:
        break