算法

1.查找最小的K个元素：利用容器来实现。如果容器中数的个数小于K时，直接插入新数。若容器存满K个数，找出K个数中最大值，当新的数据到来时，比较最大值和新数之间的大小，当新数小于最大值时，删除最大主，并且插入新数。若新数大于最大值，则抛弃新数，直接进行下一个新数的比较。

容器中K个数要找出最大值，用二叉树中的最大堆可以实现。实际中，用红黑树实现容器比较简单。红黑树通过把结点分为红、黑两种颜色并根据一些规则确保树是平衡的，从而保证在红黑树中查找、删除和插入操作都只需要O(logk)。在STL中set和multiset都是基于红黑树实现的。

void FindKLeastNumbers
(
      const vector<int>& data,               // a vector of data
      IntHeap& leastNumbers,                 // k least numbers, output
      unsigned int k                              
)
{
      leastNumbers.clear();

      if(k == 0 || data.size() < k)
            return;

      vector<int>::const_iterator iter = data.begin();
      for(; iter != data.end(); ++ iter)
      {
            // if less than k numbers was inserted into leastNumbers
            if((leastNumbers.size()) < k)
                  leastNumbers.insert(*iter);

            // leastNumbers contains k numbers and it's full now
            else
            {
                  // first number in leastNumbers is the greatest one
                  IntHeap::iterator iterFirst = leastNumbers.begin();

                  // if is less than the previous greatest number 
                  if(*iter < *(leastNumbers.begin()))
                  {
                        // replace the previous greatest number
                        leastNumbers.erase(iterFirst);
                        leastNumbers.insert(*iter);
                  }
            }
      }
}

2.两个栈实现1个队列

template<typename T> class CQueue

{

public:

CQueue(){};

~CQueue(){};

void appendTail(const &node);

void deleteHead();

private:

T->m_stack1;

T->m_statck2;

};

template<typename T> void CQueue<T>::appendTail(const T&element)

{

m_statck1.push(element);

}

template<typename T> void CQueue<T>::deleteHead()

{

if(m_statck2.size<=0)

{

while(m_stack1.size()>0)

{

T &data=m_stack1.pop();

m_stack1.pop();

m_stack.pusj(data);

}

}

assert(m_stack2.size()>0)

m_stack2.pop;

}

3.从尾到投输出链表的节点