Given a string and a non-negative int n, we’ll say that the front of the string is the first 3 chars, or whatever is there if the string is less than length 3. Return n copies of the front;
front_times(‘Chocolate’, 2) → ‘ChoCho’
front_times(‘Chocolate’, 3) → ‘ChoChoCho’
front_times(‘Abc’, 3) → ‘AbcAbcAbc’
def front_times(str, n):if len(str) < 3:return str * nelse:return str[:3] * n
Expected Run
front_times(‘Chocolate’, 2) → ‘ChoCho’ ‘ChoCho’ OK
front_times(‘Chocolate’, 3) → ‘ChoChoCho’ ‘ChoChoCho’ OK
front_times(‘Abc’, 3) → ‘AbcAbcAbc’ ‘AbcAbcAbc’ OK
front_times(‘Ab’, 4) → ‘AbAbAbAb’ ‘AbAbAbAb’ OK
front_times(‘A’, 4) → ‘AAAA’ ‘AAAA’ OK
front_times(‘’, 4) → ‘’ ‘’ OK
front_times(‘Abc’, 0) → ‘’ ‘’ OK
All Correct
def front_times(str, n):front_len = 3if front_len > len(str):front_len = len(str)front = str[:front_len]result = ""for i in range(n):result = result + frontreturn result
下一篇:《各版本机器人部署》