2014美团网校园招聘面试总结（华科）

2014年9月27日

程序主要是二叉树方面的，第二面被鄙视了。下面对二叉树的面试题做个总结。

#include <iostream>
#include <deque>
#include <vector>
using namespace std;

struct BinaryTreeNode
{
	int m_nVal;
	BinaryTreeNode *m_pLeft;
	BinaryTreeNode *m_pRight;
};

/*
	根据前序遍历和中序遍历构造二叉树
*/
BinaryTreeNode* ConstructBinaryTreeCore(int *startPreOrder, int *endPreOrder, int *startInOrder, int *endInOrder)
{
	BinaryTreeNode *pRoot = new BinaryTreeNode;

	pRoot->m_nVal = startPreOrder[0];

	pRoot->m_pLeft = pRoot->m_pRight = NULL;

	if (startPreOrder == endPreOrder)
	{
		if (startInOrder == endInOrder && *startInOrder == *endInOrder)
		{
			return pRoot;
		}
		else
		{
			return NULL;
		}
	}

	int *current = startInOrder;

	while (*current != startPreOrder[0])
	{
		++current;
	}

	int len = current - startInOrder;

	if (len > 0)
	{
		pRoot->m_pLeft = ConstructBinaryTreeCore(startPreOrder+1, startPreOrder+len, startInOrder, current-1);
	}

	if ((endInOrder-current) > 0)
	{
		pRoot->m_pRight = ConstructBinaryTreeCore(startPreOrder+len+1, endPreOrder, current+1, endInOrder);
	}

	return pRoot;
}

BinaryTreeNode* ConstructBinaryTree(int preOrder[], int inOrder[], int n)
{
	if (preOrder==NULL || inOrder==NULL || n<=0)
	{
		return NULL;
	}

	return ConstructBinaryTreeCore(preOrder, preOrder+n-1, inOrder, inOrder+n-1);
}

/*
	层次遍历二叉树
*/
void PrintByLevel(BinaryTreeNode *pRoot)
{
	if (NULL == pRoot)
	{
		return ;
	}

	deque<BinaryTreeNode *> dq;
	dq.push_back(pRoot);

	while (dq.size() > 0)
	{
		BinaryTreeNode *pNode = dq.front();
		dq.pop_front();
		cout<<pNode->m_nVal<<endl;

		if (pNode->m_pLeft != NULL)
		{
			dq.push_back(pNode->m_pLeft);
		}

		if (pNode->m_pRight != NULL)
		{
			dq.push_back(pNode->m_pRight);
		}
	}
}

/*
	二叉树的映像
*/
void Mirror(BinaryTreeNode *pRoot)
{
	if (pRoot == NULL || (pRoot->m_pLeft==NULL && pRoot->m_pRight==NULL))
	{
		return ;
	}

	BinaryTreeNode *pNode = pRoot->m_pLeft;
	pRoot->m_pLeft = pRoot->m_pRight;
	pRoot->m_pRight = pNode;

	if (pRoot->m_pLeft != NULL)
	{
		Mirror(pRoot->m_pLeft);
	}

	if (pRoot->m_pRight != NULL)
	{
		Mirror(pRoot->m_pRight);
	}
}

/*
	打印二叉树中从根节点到叶子节点的节点值之和等于k的所有路径
*/
void FindPathCore(BinaryTreeNode *pRoot, int sum, vector<int> path, int &curSum)
{
	curSum += pRoot->m_nVal;
	path.push_back(pRoot->m_nVal);

	bool isLeaf = (pRoot->m_pLeft == NULL && NULL == pRoot->m_pRight);

	if (isLeaf && curSum == sum)
	{
		cout<<"The Path is:"<<endl;
		for (vector<int>::iterator i=path.begin(); i!=path.end(); ++i)
		{
			cout<<*i<<endl;
		}
		cout<<endl;
	}

	if (pRoot->m_pLeft != NULL)
	{
		FindPathCore(pRoot->m_pLeft, sum, path, curSum);
	}

	if (pRoot->m_pRight != NULL)
	{
		FindPathCore(pRoot->m_pRight, sum, path, curSum);
	}

	curSum -= pRoot->m_nVal;
	path.pop_back();
}

void FindPath(BinaryTreeNode *pRoot, int sum)
{
	if (NULL == pRoot)
	{
		return ;
	}

	vector<int> path;
	int curSum = 0;
	FindPathCore(pRoot, sum, path, curSum);
}

int main()
{
	int preOrder[] = {1, 2, 5, 4, 6};
	int inOrder[] = {5, 2, 4, 1, 6};

	BinaryTreeNode *pRoot = ConstructBinaryTree(preOrder, inOrder, 5);
	PrintByLevel(pRoot);

	cout<<endl;
	Mirror(pRoot);
	PrintByLevel(pRoot);


	FindPath(pRoot, 7);

	system("pause");
	return 0;
}

翻转单词顺序

#include <iostream>
using namespace std;

void swap(char *a, char *b)
{
	char ch = *a;
	*a = *b;
	*b = ch;
}

void Reverse(char *pBegin, char *pEnd)
{
	if (pBegin==NULL || NULL==pEnd)
	{
		return ;
	}

	while (pBegin < pEnd)
	{
		swap(pBegin++, pEnd--);
	}
}

char* ReverseSentence(char *pStr)
{
	if (pStr == NULL)
	{
		return NULL;
	}

	char *pBegin = pStr;
	char *pEnd = pStr;

	while (*pEnd != '\0')
	{
		pEnd++;
	}
	pEnd--;

	Reverse(pBegin, pEnd);

	pBegin = pEnd = pStr;
	while (*pBegin != '\0')
	{
		if (*pBegin == ' ')
		{
			pBegin++;
			pEnd++;
		}
		else if (*pEnd == ' ' || *pEnd == '\0')
		{
			Reverse(pBegin, --pEnd);
			pBegin = ++pEnd;
		}
		else
		{
			pEnd++;
		}
	}

	return pStr;
}

int main()
{
	char pStr[] = "I am a    huster.";

	char *str = ReverseSentence(pStr);
	cout<<str<<endl;

	system("pause");
	return 0;
}