算法题记录

2025年8月6日

number_to_letter = {chr(96 + i): i for i in range(1, 27)}
def fn(str1,str2,number_to_letter):
    steps = []
    length = len(str1)
    for i in range(len(str2[:length])+1):
      
        # print("---")
        step = 0
        # 第一个循环是取出所有子串
        str2_ = str2[i:i+length]
        if length > len(str2_):
            break
        print(str1,str2_)
        
            
        for i in range(len(str1)):
            # 第二个循环是计算每一个子串的距离
            
            left = str1[i]
            right = str2_[i]
            diff = abs(number_to_letter[left] - number_to_letter[right])
            if diff >= 13:
              diff = 26-diff
            step+=diff
            # print("diff",diff)
        steps.append(step)
        # print("step",step)
    return steps

str1 = "ddhnayalcs"
str2 = "tffivyioexdqifmeadcz"
steps = fn(str1,str2,number_to_letter)
min(steps)