(LeetCode)两个队列来实现一个栈
2016-03-02 20:22
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原题例如以下:
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.
Notes:
You must use only standard operations of a queue -- which means only
and
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).
基本思路是:如果有两个队列Q1和Q2,当二者都为空时。入栈操作能够用入队操作来模拟,能够随便选一个空队列,如果选Q1进行入栈操作。如今如果a,b,c依次入栈了(即依次进入队列Q1)。这时如果想模拟出栈操作,则须要将c出栈。由于在栈顶。这时候能够考虑用空队列Q2,将a,b依次从Q1中出队,而后进入队列Q2,将Q1的最后一个元素c出队就可以。此时Q1变为了空队列。Q2中有两个元素,队头元素为a。队尾元素为b。接下来如果再运行入栈操作,则须要将元素进入到Q1和Q2中的非空队列,即进入Q2队列,出栈的话,就跟前面的一样。将Q2除最后一个元素外所有出队,并依次进入队列Q1,再将Q2的最后一个元素出队就可以。
Java实现代码例如以下:
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.
Notes:
You must use only standard operations of a queue -- which means only
push to back,
peek/pop from front,
size,
and
is emptyoperations 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).
基本思路是:如果有两个队列Q1和Q2,当二者都为空时。入栈操作能够用入队操作来模拟,能够随便选一个空队列,如果选Q1进行入栈操作。如今如果a,b,c依次入栈了(即依次进入队列Q1)。这时如果想模拟出栈操作,则须要将c出栈。由于在栈顶。这时候能够考虑用空队列Q2,将a,b依次从Q1中出队,而后进入队列Q2,将Q1的最后一个元素c出队就可以。此时Q1变为了空队列。Q2中有两个元素,队头元素为a。队尾元素为b。接下来如果再运行入栈操作,则须要将元素进入到Q1和Q2中的非空队列,即进入Q2队列,出栈的话,就跟前面的一样。将Q2除最后一个元素外所有出队,并依次进入队列Q1,再将Q2的最后一个元素出队就可以。
Java实现代码例如以下:
class MyStack { LinkedList<Integer> queue1 = new LinkedList<Integer>(); LinkedList<Integer> queue2 = new LinkedList<Integer>(); // Push element x onto stack. public void push(int x) { if(queue1.size()==0&&queue2.size()==0) { queue1.offer(x); } else if(queue1.size()==0) { queue2.offer(x); } else { queue1.offer(x); } } // Removes the element on top of the stack. public void pop() { if(queue1.size()!= 0) { int length = queue1.size(); for(int i =0;i<length-1;i++) { queue2.offer(queue1.poll()); } queue1.poll(); } else { int length = queue2.size(); for(int i =0;i<length-1;i++) { queue1.offer(queue2.poll()); } queue2.poll(); } } // Get the top element. public int top() { if(queue1.size()!= 0) { int length = queue1.size(); for(int i =0;i<length-1;i++) { queue2.offer(queue1.poll()); } int result = queue1.element(); queue2.offer(queue1.poll()); return result; } else { int length = queue2.size(); for(int i =0;i<length-1;i++) { queue1.offer(queue2.poll()); } int result = queue2.element(); queue1.offer(queue2.poll()); return result; } } // Return whether the stack is empty. public boolean empty() { return queue1.size()==0&&queue2.size()==0; } }
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