题解 | #计算字符串的编辑距离#

动态规划求解。

while True:
    try:
        s1 = input().strip()
        s2 = input().strip()
        m, n = len(s1), len(s2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m+1):
            dp[i][0] = i
        for i in range(1, n+1):
            dp[0][i] = i
        for i in range(1, m+1):
            for j in range(1, n+1):
                if s1[i-1] == s2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j]+1, dp[i][j-1]+1, dp[i-1][j-1]+1)
        print(dp[-1][-1])
    except:
        break