数据结构和算法-栈

Posted 2021-12-28 zlone

栈可以分为

• 顺序栈: 数组实现
• 链式栈: 链表实现

空间复杂度

栈的空间复杂度:
有一个n个元素的栈, 在入栈和出栈过程中, 只需要存储一个临时变量存储空间, 所以空间复杂度是O(1)

并不是说栈有n个元素, 空间复杂度就是O(n), 而是指除了原本的空间外, 算法需要的额外空间

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栈要满足后进先出(LIFO)的特性, 栈有以下几种方法

• 判断为空isEmpty
• 入栈push
• 出栈pop
• 返回栈顶元素peek
• 返回栈大小size
• 是否是空isEmpty

以下是使用列表来模拟栈的操作

# coding:utf-8

class Stack(object):
    """模拟栈"""
    def __init__(self):
        self.items = []

    def isEmpty(self):
        return self.items == []

    def push(self, item):
        self.items.append(item)

    def pop(self):
        if self.isEmpty():
            raise Exception('the stack is empty')
        return self.items.pop()

    def peek(self):
        if self.isEmpty():
            raise Exception('the stack is empty')
        return self.items[-1]

    def size(self):
        return len(self.items)


if __name__ == '__main__':
    s = Stack()
    s.push('t')
    s.push('e')
    s.push('s')

    assert s.pop() == 's'
    assert s.peek() == 'e'
    assert s.pop() == 'e'
    assert s.pop() == 't'

应用

1. 符号匹配

实现括号匹配来区分符号是否平衡. 如果是开始符号如(, , [那么就压入栈, 如果碰到结束符号, 则从栈顶弹出元素

class Stack(object):
    def __init__(self):
        self.stack = []

    def push(self, item):
        self.stack.append(item)

    def pop(self):
        return self.stack.pop()

    def isEmpty(self):
        return self.stack == []


basic = 
    ')': '(',
    ']': '[',
    '': '',



def test(string):
    s = Stack()
    first = basic.values()
    last = basic.keys()
    for i in string:
        if i in first:
            s.push(i)
        elif i in last:
            if s.pop() == basic[i]:
                continue
            else:
                return False
        else:
            continue
    if s.isEmpty():
        return True
    return False


if __name__ == '__main__':
    assert test('[hello ( world  === )]') == True
    assert test('kk(dsfd)ll[') == False

2. 求二进制

如数字6(110), 分别用2除6, 求余数, 最后余数反转就是110

class Stack(object):
    def __init__(self):
        self.stack = []

    def push(self, item):
        self.stack.append(item)

    def pop(self):
        return self.stack.pop()

    def isEmpty(self):
        return self.stack == []


def binary(num):
    s = Stack()
    while num > 0:
        n = num % 2
        s.push(n)
        num = num // 2
    res = ""
    while not s.isEmpty():
        res += str(s.pop())
    return res


if __name__ == "__main__":
    assert binary(5) == '101'
    assert binary(8) == '1000'
    assert binary(9) == '1001'