LeetCode 225. Implement Stack using Queues
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问题描述:
Implement the following operations of a stack using queues.
- push(x) -- Push element x onto stack.
- pop() -- Removes the element on top of the stack.
- top() -- Get the top element.
- empty() -- Return whether the stack is empty.
- You must use only standard operations of a queue -- which means only
push to back
,peek/pop from front
,size
, andis empty
operations are valid. - Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
- You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
分析:用先进先出的队列实现后进先出的栈。
进栈:这里选取queue1进栈,直接queue.push(x)
出栈:即删除最后一个进入queue1的的元素,若queue1大小为1,则直接pop,若大于1,则把先进queue1的几个元素移入queue2中,至queue1只剩一个元素,然后删除钙元素,删除该元素后要记住:将queue2中的元素依次移入queue1
取栈顶元素:同出栈差不多,但有一点要注意:queue1只剩1个元素的时候,它就是栈顶元素,但同时,还是要把它移入queue2,然后在把queue2中的元素依次移入queue1,这么做是为了保证次序不变!!
AC代码如下:
class Stack {public: queue<int>queue1; queue<int>queue2; // Push element x onto stack. void push(int x) { queue1.push(x); } // Removes the element on top of the stack. void pop() { if(queue1.size() == 1) queue1.pop(); else { while(queue1.size() > 1) { int temp = queue1.front(); queue1.pop(); queue2.push(temp); } queue1.pop(); //将队列2中的元素放回队列1,保证队列1中总有元素 while(queue2.size() > 0) { int temp = queue2.front(); queue2.pop(); queue1.push(temp); } } } // Get the top element. int top() { if(queue1.size() == 1) return queue1.front(); else { while(queue1.size() > 1) { int temp = queue1.front(); queue1.pop(); queue2.push(temp); } int res = queue1.front();//找到栈定元素 //找到栈定元素之后,一定要把栈定元素放到队列2中 queue1.pop(); queue2.push(res); //将队列2中的元素放回队列1,保证队列1中总有元素 while(queue2.size() > 0) { int temp = queue2.front(); queue2.pop(); queue1.push(temp); } return res; } } // Return whether the stack is empty. bool empty() { return queue1.empty() && queue2.empty(); }};
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