题解 | 小红的字符串修改

s1 = input()
s2 = input()
l1, l2 = len(s1), len(s2)
# 初始化一个很大的数
min_times = 1e6
# 转为ASCII码列表，便于计算
s1_num = [ord(i) for i in s1]
s2_num = [ord(i) for i in s2]

for i in range(l2):
    # 条件判断，防止下标越界
    if i + l1 <= l2:
        times = 0
        for j, n in enumerate(s1_num):
            # 两个方向计算最小转换步数
            times += min(abs(s2_num[i + j] - s1_num[j]), 26-abs(s2_num[i + j] - s1_num[j]))
        min_times = min(min_times, times)
    else:
        break
print(min_times)

类似于指针，每次截取一段直接计算所有可能取最小值