【剑指offer】07与09
1.面试题07. 重建二叉树
输入某二叉树的前序遍历和中序遍历的结果,请重建该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
例如,给出
前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树:
3
/ \
9 20
/ \
15 7
限制:0 <= 节点个数 <= 5000
先序遍历,先根结点,后左子树,右子树
中序遍历,先左子树,后根结点,右子树
所以先序遍历第一个为根结点,从中序遍历中搜索根结点位置,根结点前为左子树,根结点后为右子树,分别对左子树右子树进行递归
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if(preorder.size()==0||inorder.size()==0) return 0;
return build(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
}
TreeNode* build(vector<int>& preorder,int p1,int q1,vector<int>& inorder,int p2,int q2)
{
TreeNode* root=new TreeNode(preorder[p1]);
int i,lc,rc;
i=p2;
while(preorder[p1]!=inorder[i]) i++;
lc=i-p2;rc=q2-i;
if(lc>0) root->left=build(preorder,p1+1,p1+lc,inorder,p2,i-1);
if(rc>0) root->right=build(preorder,p1+lc+1,q1,inorder,i+1,q2);
return root;
}
};
2.面试题09. 用两个栈实现队列
用两个栈实现一个队列。队列的声明如下,请实现它的两个函数 appendTail 和 deleteHead ,分别完成在队列尾部插入整数和在队列头部删除整数的功能。(若队列中没有元素,deleteHead 操作返回 -1 )
示例 1:
输入:
["CQueue","appendTail","deleteHead","deleteHead"]
[[],[3],[],[]]
输出:[null,null,3,-1]
示例 2:
输入:
["CQueue","deleteHead","appendTail","appendTail","deleteHead","deleteHead"]
[[],[],[5],[2],[],[]]
输出:[null,-1,null,null,5,2]
提示:
1 <= values <= 10000
最多会对 appendTail、deleteHead 进行 10000 次调用
class CQueue {
public:
CQueue() {
}
std::stack<int> stk1;
std::stack<int> stk2;
void appendTail(int value) {
stk1.push(value);
}
int deleteHead() {
if(stk1.empty() && stk2.empty()){
return -1;
}
if(stk2.empty()){
while (!stk1.empty()){
stk2.push(stk1.top());
stk1.pop();
}
}
int d = stk2.top();
stk2.pop();
return d;
}
};
/**
* Your CQueue object will be instantiated and called as such:
* CQueue* obj = new CQueue();
* obj->appendTail(value);
* int param_2 = obj->deleteHead();
*/