Article Directory
Foreword
Code is just some ideas, some basis for comparison not be verified, if the problem or have better wording, welcome advice.
Byte beating
A topic
Reverse binary tree
void Fun(Tree * node){
if(node!=nullptr){
Fun(node->left);
Fun(node->right);
Tree * temp = node->left;
node->left = node->right;
node->right = temp;
}
}
US Mission
A topic
In order to achieve the traversal cycle
following the cycle is panacea:
void Fun(Tree* node){
stack<Tree *> trees;
stack<bool> signs;
trees.push(node);
signs.push(false);
while(trees.size()>0){
Tree *tree = trees.top();
bool sign = signs.top();
trees.pop();
signs.pop();
if(tree!=nullptr){
if(sign){
cout<<tree->val<<endl;
}else{
//前序遍历
//trees.push(tree->right);
//trees.push(tree->left);
//trees.push(tree);
//signs.push(false);
//signs.push(false);
//signs.push(true);
//中序遍历
trees.push(tree->right);
trees.push(tree);
trees.push(tree->left);
signs.push(false);
signs.push(true);
signs.push(false);
//后序遍历
//trees.push(tree->right);
//trees.push(tree->left);
//trees.push(tree);
//signs.push(false);
//signs.push(false);
//signs.push(true);
}
}
}
}
The results of the interviewer to the sentence, this method you need more of a bool type of memory, can not do without the extra space to achieve it?
//中序遍历
void Fun(Tree *node){
stack<Tree*> trees;
trees.push(node);
node = node->left;
while(trees.size()>0){
if(node!=nullptr){
trees.push(node);
node = node->left;
}else{
cout<<trees.top()->val<<endl;
node = trees.top();
trees.pop();
}
}
}
So preorder traversal of it?
//前序遍历
void Fun(Tree *node){
stack<Tree*> trees;
trees.push(node);
while(trees.size()>0){
Tree *tree = trees.top();
trees.pop();
if(tree!=nullptr){
cout<<tree->val<<endl;
trees.push(tree->right);
trees.push(tree->left);
}
}
}
So postorder it? Postorder seemingly not, ha ha, go panacea it.
Ape counseling
A topic
ASC has a array, moved to the right of the k bits:
1 2 3 4 k=0
4 1 2 3 k=1
3 4 1 2 k=2
...
Now we know a list, returns k.
//最简单的做法，找到那个不是升降排序的节点，返回即可
int Fun(vector<int> nums){
for(int i=1;i<nums.size();++i){
if(nums[i-1]>nums[i]){
return i+1;
}
}
return 0;
}
Then the interviewer said, the complexity of the algorithm is O (N), there is no easier. There, look for a half:
int Fun(vector<int> nums){
//判断数组是否为空，且是否k=0;
if(nums.size() == 0 || nums[0] < nums.back())
return 0;
int low=0;
int high=nums.size()-1;
while(low<high){
int mid = (low + high) / 2;
if(mid == low)
break;
if(nums[mid] > nums[low])
low = mid;
else if(nums[mid] < nums[high])
high = mid;
}
return low;
}
Topic two
To the root node of a binary tree, to the value of two nodes, the minimum required to find their common parent. For example, the following tree:
To D, F, the minimum common parent A
to H, E, minimum common parent B
struct Tree{
int val;
Tree *left;
Tree *right;
};
Tree * Fun(Tree* node,int A,int B){
if(node!=nullptr){
if(node->val == A || node->val == B)
return node;
Tree *left = Fun(node->left,A,B);
Tree *right= Fun(node->right,A,B);
if(left!=nullptr&&right!=nullptr)
return node;
else if(left!=nullptr)
return left;
else if(right!=nullptr)
return right;
}
return node;
}
Topic three
An unordered list to request the list into odd and even chain, while the parity Yaoan list in ascending order.
struct List{
int val;
List * next;
};
List* Insert(List* root,List* node){
if(root == nullptr){
node->next = nullptr;
return node;
}else if(node->val<root->val){
node->next = root;
return node;
}
root->next = Insert(root->next,node);
return root;
}
void Fun(List *root){
List * odd = nullptr;
List * even = nullptr;
while(root!=nullptr){
List * node = root;
root = root->next;
if(node->val%2==1){
odd = Insert(odd,node);
}else{
even = Insert(even, node);
}
}
}
Tencent
first question
How to generate a random number between 0 and 100 and 100 can not be repeated.
void swap(int &x,int &y){
int temp = x;
x = y;
y = temp;
}
void Fun(){
vector<int> nums(100);
//赋值
for(int i=0;i<100;++i)
nums[i] = i;
//每次随机取一个数和最后面的数字进行交换
for(int i=100;i>0;--i){
int r = rand()%(i+1);
swap(nums[r],nums[i-1]);
}
}
The second question
Multiplying large numbers (multiplied by the array)
vector<int> Fun(vector<int> num1,vector<int> num2){
vector<int> num(num1.size()+num2.size());
//先不考虑进位，把每位乘得的结果加到对应的位置上去
for(int i=(int)num1.size()-1;i>=0;--i){
int index = (int)num1.size()-i-1;
for(int j=(int)num2.size()-1;j>=0;--j,++index){
num[index]+=num1[i]*num2[j];
}
}
//准备进位
for(int i=0,sign=0;i<num.size();++i){
num[i]+=sign;
sign = num[i]/10;
num[i] = num[i]%10;
}
//去掉多余的位数
while(num.back()==0)
num.pop_back();
//反转
reverse(num.begin(), num.end());
return num;
}
quick worker
first question
Binary tree traversal performed by the hierarchy.
void Fun(Tree * root){
vector<Tree *> trees;
trees.push_back(root);
int index = 0;
while(index<trees.size()){
if(trees[index]!=nullptr){
trees.push_back(trees[index]->left);
trees.push_back(trees[index]->right);
}
++index;
}
for(int i=0;i<trees.size();++i){
if(trees[i] != nullptr){
cout<<trees[i]->val<<" ";
}
}
}
