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  • Leetcode——Queue队列

    Queue 队列

    性质:先入先出FIFO

    在FIFO数据结构中,将首先处理添加到其中的第一个元素。队列是典型的 FIFO 数据结构。

    • 插入(insert)操作也称作入队(enqueue),新元素始终被添加在队列的末尾。
    • 删除(delete)操作也被称为出队(dequeue)。 你只能移除第一个元素。

    队列的实现(python)——列表list实现

    class Queue:
        def __init__(self):
            self.items=[]
        def isEmpty(self):
            return self.items==[]
        def enqueue(self,item):
            self.items.insert(0,item)
        def dequeue(self):
            return self.items.pop()
        def  size(self):
            return len(self.items)
    
    

    循环队列的实现

    代码

    class MyCircularQueue:
        def __init__(self, k: int):
            """
            Initialize your data structure here. Set the size of the queue to be k.
            """
            self.item = [-1 for _ in range(k)]
            self.maxsize = k
            self.size = 0
            self.font = -1
            self.rear = 0
        def enQueue(self, value: int) -> bool:
            """
            Insert an element into the circular queue. Return true if the operation is successful.
            """
            if self.isFull():
                return False
            else:
                self.item[self.rear] = value
                self.rear = (self.rear+1)%self.maxsize
                self.size += 1
                return True
        def deQueue(self) -> bool:
            """
            Delete an element from the circular queue. Return true if the operation is successful.
            """
            if self.isEmpty():
                return False
            else:
                self.font = (self.font+1)%self.maxsize
                self.size -= 1
                return True
        def Front(self) -> int:
            """
            Get the front item from the queue.
            """
            if not self.isEmpty():
                return self.item[self.font+1]
            else:
                return -1
        def Rear(self) -> int:
            """
            Get the last item from the queue.
            """
            if not self.isEmpty():
                return self.item[self.rear-1]
            else:
                return -1
        def isEmpty(self) -> bool:
            """
            Checks whether the circular queue is empty or not.
            """
            return self.size == 0
        def isFull(self) -> bool:
            """
            Checks whether the circular queue is full or not.
            """
            return self.size == self.maxsize       
    # Your MyCircularQueue object will be instantiated and called as such:
    # obj = MyCircularQueue(k)
    # param_1 = obj.enQueue(value)
    # param_2 = obj.deQueue()
    # param_3 = obj.Front()
    # param_4 = obj.Rear()
    # param_5 = obj.isEmpty()
    # param_6 = obj.isFull()
    

    代码解释

    • self.font:代表循环队列列首的前一个位置,也就是self.font=real_font-1
    • self.rear:代表循环队列列尾的后一个位置,也就是self.rear=real_rear+1
    • self.size:判断队列的空满,当然也可以用下列方式判断队列的空满
      • (real_rear+1)%maxsize=real_font,也就是(self.rear)%self.maxsize==self.font+1:再添加一个元素之后,队首self.font+1和队尾(self.rear)%self.maxsize重合
      • (real_font+1)%maxsize=real_rear,也就是(self.font+2)%self.maxsize==self.rear-1:再删除一个元素之后,队首(self.font+2)%self.maxsize和队尾self.rear-1重合
        图示

    队列Queue与广度优先搜索BFS

    #->功能:广度优先搜索算法
    #->传入参数:g(图),start(起始节点)
    def bfs(g,start):
        #当前顶点的distance predecessor的设置
        start.setDistance(0)
        start.setPred(None)
        #新建队列,存储顶点
        vertQueue = Queue()
        #当前节点进队
        vertQueue.enqueue(start)
        while (vertQueue.size() > 0):  
            currentVert = vertQueue.dequeue()  #取队首为当前顶点
            for nbr in currentVert.getConnections()[0]: #遍历当前顶点的邻接顶点
                if nbr.getColor() == 'white':  #邻接顶点白色(未遍历)
                    nbr.setColor('gray')     #遍历过了
                    nbr.setDistance(currentVert.getDistance()+1) #当前顶点的距离加1作为该遍历节点的距离
                    nbr.setPred(currentVert)  #设置前驱
                    vertQueue.enqueue(nbr)   #邻接节点入队
            currentVert.setColor('black')  #当前节点全部遍历
    
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  • 原文地址:https://www.cnblogs.com/zjx-pku/p/13173638.html
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