爱上算法：每日算法（24-2月4号）

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🌟坚持每日刷算法，😃将其变为习惯🤛让我们一起坚持吧💪

232. 用栈实现队列

思路

首先应该先明确队列是先进先出,

而栈是先进后出,而如果想用栈实现队列,就可以尝试用两个栈

进栈和出栈

• 进栈模拟入队列
• 出栈模拟先出队列

画图如下

Code

Java

class MyQueue {Stack<Integer> stIn;Stack<Integer> stOut;public MyQueue() {stIn = new Stack<>();stOut = new Stack<>();}public void push(int x) {stIn.push(x);}public int pop() {if(stOut.isEmpty()){while(!stIn.isEmpty()){stOut.push(stIn.pop());} }return stOut.pop();}public int peek() {int res = this.pop();stOut.push(res);return res;}public boolean empty() {return stOut.isEmpty()&&stIn.isEmpty();}
}/*** 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();*/

C++

class MyQueue {
public:stack<int>	 stIn;stack<int> stOut;MyQueue() {}void push(int x) {stIn.push(x);}int pop() {if(stOut.empty()){while(!stIn.empty()){stOut.push(stIn.top());stIn.pop();}}int result = stOut.top();stOut.pop();return result;}int peek() {int res = this->pop();stOut.push(res);return res;}bool empty() {return stIn.empty()&&stOut.empty();}
};/*** 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();* bool param_4 = obj.empty();*/

复杂度

时间复杂度: push和empty为O(1), pop和peek为O(n)

空间复杂度: O(n)

225. 用队列实现栈

思路

队列是先进先出原则,

而栈是先进后出原则

因此,可以使用两个队列来实现栈

可以使用一个队列来实现栈

满足先进后出的方法就是; 入队列之后,就将这个数放到队首

Code

C++

class MyStack {
public:	queue<int> que;MyStack() {}void push(int x) {que.push(x);}int pop() {int size = que.size();size--;while(size--){que.push(que.front());que.pop();}int res = que.front();que.pop();return res;}int top() {return que.back();}bool empty() {return que.empty();}
};/*** 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();* bool param_4 = obj->empty();*/

Java

class MyStack {Queue<Integer> que = new LinkedList<>();public MyStack() {}public void push(int x) {que.add(x);}public int pop() {rePosition();return que.poll();}public int top() {rePosition();int res = que.poll();que.add(res);return res;}public boolean empty() {return que.isEmpty();}public void rePosition(){int size = que.size();size--; // 不包括刚刚添加的数while(size-- > 0){que.add(que.poll());}}
}/*** 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();*/

复杂度

• 时间复杂度: pop为O(n)，其他为O(1)
• 空间复杂度: O(n)

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