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[LeetCode] Implement Queue using Stacks
2015-11-21 00:57:52 来源: 作者: 【 】 浏览:2
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Implement Queue using Stacks

Implement the following operations of a queue using stacks.

  • push(x) -- Push element x to the back of queue.
  • pop() -- Removes the element from in front of queue.
  • peek() -- Get the front element.
  • empty() -- Return whether the queue is empty.

    Notes:

    • You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid.
    • Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
    • You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

      解题思路:

      用两个栈来模拟队列。其中一个栈做缓存之用,另外一个栈只接受。push操作,直接放入接收之栈中。pop操作,若缓存栈存在数据,则从缓存栈中弹出。若缓存栈为空,则将接收栈的数据弹出并插入到缓存栈中,再从缓存栈中弹出。peek操作与pop操作类似。当且仅当两个栈都为空时,empty才返回true。

      ?

      class Queue {
      public:
          // Push element x to the back of queue.
          void push(int x) {
              stacks[0].push(x);
          }
      
          // Removes the element from in front of queue.
          void pop(void) {
              if(stacks[1].empty()){
                  while(!stacks[0].empty()){
                      stacks[1].push(stacks[0].top());
                      stacks[0].pop();
                  }
              }
              stacks[1].pop();
          }
      
          // Get the front element.
          int peek(void) {
              if(stacks[1].empty()){
                  while(!stacks[0].empty()){
                      stacks[1].push(stacks[0].top());
                      stacks[0].pop();
                  }
              }
              return stacks[1].top();
          }
      
          // Return whether the queue is empty.
          bool empty(void) {
              return stacks[0].empty() && stacks[1].empty();
          }
          
      private:
          stack
            
              stacks[2];
      };
            


      ?

      版权声明:本文为博主原创文章,未经博主允许不得转载。

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