leetcode刷题python之多线程 按序打印

import threading

class Foo:
    def __init__(self):
        self.c = threading.Condition()
        self.t = 0


    def first(self, printFirst: 'Callable[[], None]') -> None:
        
        # printFirst() outputs "first". Do not change or remove this line.
        self.res(0, printFirst)

    def second(self, printSecond: 'Callable[[], None]') -> None:
        
        # printSecond() outputs "second". Do not change or remove this line.
        self.res(1, printSecond)#里面有没有括号

    def third(self, printThird: 'Callable[[], None]') -> None:
        
        # printThird() outputs "third". Do not change or remove this line.
        self.res(2, printThird)
        
    def res(self, val: int, func: 'Callable[[], None]') -> None:
        with self.c:
            self.c.wait_for(lambda: val == self.t)
            func()
            self.t += 1
            self.c.notify_all()

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