线性表 - 栈和队列

后进先出 LIFO

两种实现方式

  • 使用数组实现的叫静态栈
  • 使用链表实现的叫动态栈

相关题目

简单难度

225. 用队列实现栈 https://leetcode.cn/problems/implement-stack-using-queues/
class MyStack {
    private Queue<Integer> q1;
    private Queue<Integer> q2;
    public MyStack() {
        q1 = new LinkedList<>();
        q2 = new LinkedList<>();
    }
   
    public void push(int x) {
        q2.offer(x);
        while (!q1.isEmpty()) {
            q2.offer(q1.poll());
        }
        Queue<Integer> temp = q1;
        q1 = q2;
        q2 = temp;
    }
   
    public int pop() {
        return q1.poll();
    }
   
    public int top() {
        return q1.peek();
    }
   
    public boolean empty() {
        return q1.isEmpty();
    }
}

中等难度

71. 简化路径 https://leetcode.cn/problems/simplify-path/
class Solution {
    public String simplifyPath(String path) {
        String[] names = path.split("/");
        Deque<String> stack = new ArrayDeque<>();
        for (String name : names) {
            if ("..".equals(name)) {
                if (!stack.isEmpty()) {
                    stack.pollLast();
                }
            } else if (name.length() > 0 && !".".equals(name)) {
                stack.offerLast(name);
            }
        }
        StringBuffer ans = new StringBuffer();
        if (stack.isEmpty()) {
            ans.append('/');
        } else {
            while (!stack.isEmpty()) {
                ans.append('/');
                ans.append(stack.pollFirst());
            }
        }
        return ans.toString();
    }
}

队列

先进先出 FIFO

相关题目

简单难度

933. 最近的请求次数 https://leetcode.cn/problems/number-of-recent-calls/
class RecentCounter {
    Queue<Integer> q;
    public RecentCounter() {
        q = new LinkedList<>();
    }
   
    public int ping(int t) {
        q.offer(t);
        while (q.peek().intValue() < t - 3000) {
            q.remove();
        }
        return q.size();
    }
}

中等难度

641. 设计循环双端队列 https://leetcode.cn/problems/design-circular-deque/
class MyCircularDeque {
    private int[] elements;
    private int rear, front;
    private int capacity;
    public MyCircularDeque(int k) {
        capacity = k + 1;
        rear = front = 0;
        elements = new int[k + 1];
    }
   
    public boolean insertFront(int value) {
        if (isFull()) {
            return false;
        }
        front = (front - 1 + capacity) % capacity;
        elements[front] = value;
        return true;
    }
   
    public boolean insertLast(int value) {
        if (isFull()) {
            return false;
        }
        elements[rear] = value;
        rear = (rear + 1) % capacity;
        return true;
    }
   
    public boolean deleteFront() {
        if (isEmpty()) {
            return false;
        }
        front = (front + 1) % capacity;
        return true;
    }
   
    public boolean deleteLast() {
        if (isEmpty()) {
            return false;
        }
        rear = (rear - 1 + capacity) % capacity;
        return true;
    }
   
    public int getFront() {
        if (isEmpty()) {
            return -1;
        }
        return elements[front];
    }
   
    public int getRear() {
        if (isEmpty()) {
            return -1;
        }
        return elements[(rear - 1 + capacity) % capacity];
    }
   
    public boolean isEmpty() {
        return rear == front;
    }
   
    public boolean isFull() {
        return (rear + 1) % capacity == front;
    }
}
/**
 * Your MyCircularDeque object will be instantiated and called as such:
 * MyCircularDeque obj = new MyCircularDeque(k);
 * boolean param_1 = obj.insertFront(value);
 * boolean param_2 = obj.insertLast(value);
 * boolean param_3 = obj.deleteFront();
 * boolean param_4 = obj.deleteLast();
 * int param_5 = obj.getFront();
 * int param_6 = obj.getRear();
 * boolean param_7 = obj.isEmpty();
 * boolean param_8 = obj.isFull();
 */
posted @ 2024-01-25 22:14  白缺  阅读(4)  评论(0编辑  收藏  举报