字符串：字符串距离计算[字典]

参考 @ztxz16 的解答，改写了个简短的python代码：

@ztxz16: https://blog.nowcoder.net/n/21d889d1febd4f9998e70280b813846a

def cal(s1, s2):
    dp = [[0] * 26 for _ in range(26)]  # 在某个位置上，有dp[x1][x2]个s1[x1]和s2[x2]
    sums, res = 0, 0
    for i in range(len(s1)):
        dp[ord(s1[i]) - ord('a')][ord(s2[i]) - ord('a')] += 1
        sums += (s1[i] != s2[i])
    return sums - max([dp[i][j] - dp[i][i] for j in range(26) for i in range(26)])