代码随想录day11 Java版

232.用栈实现队列

模拟类型的题目

只要知道栈和队列的结构，用第一个栈模拟时发现能入队但出队时顺序相反了，正好利用栈能翻转输出顺序的特性，创建另一个栈用于输出

注意代码实现中，出队时如果out栈空，需要把所有in栈的元素压入out栈来维护顺序

class MyQueue {
    private Stack in;
    private Stack out;
    public MyQueue() {
        in = new Stack<>();
        out = new Stack<>();
    }
    
    public void push(int x) {
        in.push(x);
    }
    
    public int pop() {
        if (out.empty()) {
            while(!in.empty()) {
                int x = in.peek();
                out.push(x);
                in.pop();
            }
        }
        int res = out.peek();
        out.pop();
        return res;
    }
    
    public int peek() {
        if (out.empty()) {
            while(!in.empty()) {
                int x = in.peek();
                out.push(x);
                in.pop();
            }
        }
        int res = out.peek();
        //out.pop();
        return res;
    }
    
    public boolean empty() {
        return in.empty()&&out.empty() ? true:false;
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */

225. 用队列实现栈

先直接用了Java的Deque接口实现

没有特殊要求的话栈和队列都可以用Deque x = new LinkedList<>();

其中栈的方法主要有push,pop,peek,empty,队列的方法可以看下图

class MyStack {
    Deque store;
    public MyStack() {
        store = new LinkedList<>();
    }
    
    public void push(int x) {
        store.addFirst(x);
    }
    
    public int pop() {
        return store.removeFirst();
    }
    
    public int top() {
        return store.getFirst();
    }
    
    public boolean empty() {
        return store.size() == 0;
    }
}

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack obj = new MyStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * boolean param_4 = obj.empty();
 */

下面按照题目要求使用只能先入先出后入后出的Queue

offer向队尾插入，poll弹出队头

模拟入栈后发现出栈时没办法弹出队尾来实现，只能先把前面的元素都出队后才能弹出该元素

所以使用辅助队列来接收前面的元素

class MyStack {
    private Queue q;
    private Queue qq;
    public MyStack() {
        q = new LinkedList<>();
        qq = new LinkedList<>();
    }
    
    public void push(int x) {
        q.offer(x);
    }
    
    public int pop() {
        while(q.size()>1) qq.offer(q.poll());
        int res = q.poll();
        Queue tmp = q;
        q = qq;
        qq = tmp;
        return res;
    }
    
    public int top() {
        while(q.size()>1) qq.offer(q.poll());
        int res = q.peek();
        qq.offer(q.poll());
        Queue tmp = q;
        q = qq;
        qq = tmp;
        return res;
    }
    
    public boolean empty() {
        return q.isEmpty();
    }
}

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack obj = new MyStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * boolean param_4 = obj.empty();
 */

20. 有效的括号

经典用栈处理括号匹配的问题，虽然但是中缀表达式什么的没考，就直接处理三种括号不匹配的情况（左括号更多，右括号更多，左右不一样）就好了

class Solution {
    public boolean isValid(String s) {
        Deque stack = new LinkedList<>();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') stack.push(')');
            else if (s.charAt(i) == '[') stack.push(']');
            else if (s.charAt(i) == '{') stack.push('}');
            else {
                if (!stack.isEmpty() && stack.peek() == s.charAt(i)) stack.pop();
                else return false; 
            }
        }
        if (stack.isEmpty()) return true;
        else return false;
    }
}