题解 | #字符串排序#

我写的比较复杂看着。

def check_l(l: list):
    if len(l) > 1 and l[0].isdigit() and 1 <= int(l[0]) <= 1000 and int(l[0]) == (len(l)-1):
        for i in l[1:]:
            if len(i) > 100 or len(i) < 1:
                return False
    else:
        return False
    return True
    
while True:
    try:
        s = input()
        s_l.append(s)
    except EOFError:
        if check_l(s_l):
            s_l.pop(0)
            new_list = list(sorted(s_l))
            for i in new_list:
                print(i)
        break