记录
######################################### # File name :compile.sh # Author :kangkangliang # File desc : # Mail :liangkangkang@paag.com # Create time :2015-11-06 ######################################### #!/usr/bin/bash currentPath=$(pwd | sed 's$\/$\\\/$g') echo $currentPath find . name "*.h"o name "*.c"o name "*.cc"o name "*.cpp"| sed "s/^\./$currentPath/"> cscope.files cscope -bkq i cscope.files ctags -R
CentOS 7的yum源中貌似没有正常安装mysql时的mysql-sever文件，需要去官网上下载
# wget http://dev.mysql.com/get/mysql-community-release-el7-5.noarch.rpm
# rpm -ivh mysql-community-release-el7-5.noarch.rpm
# yum install mysql-community-server
成功安装之后重启mysql服务
# service mysqld restart
初次安装mysql是root账户是没有密码的
设置密码的方法
# mysql -uroot
mysql> set password for ‘root’@‘localhost’ = password('mypasswd');
mysql> exit
#include <stdio.h>
#include <stdlib.h>
typedef struct BiTree{
int val;
struct BiTree *lchild,*rchild;
}BiTree;
void Create_Sort_Tree(BiTree **t, int value)
{
if(*t == NULL){
*t = (BiTree *)malloc(sizeof(BiTree));
(*t)->val = value;
(*t)->lchild = NULL;
(*t)->rchild = NULL;
}
else{
if((*t)->val > value)
Create_Sort_Tree(&(*t)->lchild, value);
else
Create_Sort_Tree(&(*t)->rchild, value);
}
}
void lar(BiTree *t)
{
if(t == NULL)
return;
else{
lar(t->lchild);
printf("%d\t",t->val);
lar(t->rchild);
}
}
void Preorder(BiTree *t)
{
if(t)
{
printf("%d\t ", t->val);
Preorder(t->lchild);
Preorder(t->rchild);
}
}
void Inorder(BiTree *t)
{
if(t)
{
Inorder(t->lchild);
printf("%d\t ", t->val);
Inorder(t->rchild);
}
}
void Postorder(BiTree *t)
{
if(t)
{
Postorder(t->lchild);
Postorder(t->rchild);
printf("%d\t ", t->val);
}
}
/**根据前序周游和中序周游，重建一颗二叉树。
* @param pre 前序周游的结果，其中每一个字符表示一个节点的值
* @param mid 中序周游的结果，其中每一个字符表示一个节点的值
* @param n 该树节点总数。
* @return 生成的树的根节点。
*/
BiTree* buildTree(char *pre, char *mid, int n)
{
if (n==0) return NULL;
char c = pre[0];
// BiTree*node = new TreeNode(c); //This is the root node of this tree/sub tree.
BiTree *node = (BiTree*)malloc(sizeof(BiTree));
int i;
for(i=0; i<n && mid[i]!=c; i++);
int lenL = i; // the node number of the left child tree.
int lenR = n - i -1; // the node number of the rigth child tree.
//build the left child tree. The first order for thr left child tree is from
// starts from pre[1], since the first element in pre order sequence is the root
// node. The length of left tree is lenL.
if(lenL > 0) node->lchild = buildTree(&pre[1], &mid[0], lenL);
//build the right child tree. The first order stree of right child is from
//lenL + 1(where 1 stands for the root node, and lenL is the length of the
// left child tree.)
if(lenR > 0) node->rchild = buildTree(&pre[lenL+1], &mid[lenL+1], lenR);
return node;
}
BiTree* ConstructCore(int *start_pre,int *end_pre,int *start_in,int* end_in)
{
// 先序列的第一个节点为根节点
int key = start_pre[0];
BiTree *root = (BiTree*)malloc(sizeof(BiTree));
root->val = key;
root->rchild = root->lchild = NULL;
///quit condition
if(start_pre == end_pre){
// if(start_in == end_in && *start_pre == *start_in)
if(start_in == end_in)
return root;
else
printf("invalid input\n");
}
// 在中序序列中找到根节点
int *root_in = start_in;
while(root_in <= end_in && *root_in != key)
++root_in;
if(root_in == end_in && *root_in != key)
printf("invalid input\n");
int left_len = root_in - start_in;
int *left_pre_end = start_pre + left_len;
if(left_len > 0){
// create left subtree
// root->lchild = ConstructCore(start_in+1,left_pre_end,start_in,root_in-1);
root->lchild = ConstructCore(start_pre+1,left_pre_end,start_in,root_in-1);
}
if(left_len < end_pre - start_pre){
// create right subtree
root->rchild = ConstructCore(left_pre_end+1,end_pre,root_in+1,end_in);
}
return root;
}
// 根据先序和中序来构建二叉树
BiTree * Construct(int *pre,int *in,int len)
{
if(pre == NULL || in == NULL || len <= 0)
return NULL;
return ConstructCore(pre,pre+len-1,in,in+len-1);
}
int SumNode(BiTree *root)
{
int count = 0;
if(root == NULL)
return 0;
else if(root->rchild == NULL && root->lchild == NULL)
return 1;
else
count = SumNode(root->lchild)+SumNode(root->rchild);
return count;
}
unsigned int FullNodeSum(BiTree *root)
{
unsigned int count = 0;
if(root == NULL)
return 0;
else if(root->lchild == NULL && root->rchild == NULL)
return 0;
else if(root->lchild != NULL && root->rchild == NULL)
count = FullNodeSum(root->lchild);
else if(root->left == NULL && root->right != NULL)
count = FullNodeSum(root->right);
else
// 1 key code 防止只有一个结点
count = 1 + FullNodeSum(roott->left) + FullNodeSum(root->right);
return count;
}
int main(void)
{
int i;
BiTree *t = NULL;
int value[] = {5,8,14,36,21,1,3};
for(i = 0;i < 7;i++)
Create_Sort_Tree(&t,value[i]);
// lar(t);
printf("\n");
Preorder(t);
printf("\n");
// two
printf("one\n");
// int pre[8] = {1,2,4,7,3,5,6,8};
// int in[8] = {4,7,2,1,5,3,8,6};
// // char pre[] = "12473568";
// // char in[] = "47215386";
// BiTree *t2 = Construct(pre,in,8);
int preorder[] = {4,2,1,3,7,6,5,8};
int inorder[] = {1,2,3,4,5,6,7,8};
BiTree* t2 = Construct(preorder,inorder,8);
Preorder(t2);
printf("\n");
Inorder(t2);
printf("\n");
Postorder(t2);
printf("\n");
printf("leaf :\n");
int leaf = SumNode(t2);
printf("%d\n", leaf);
}
tree
/*
delete dup node
--------
if(pHead==NULL || pHead->next==NULL){
return pHead;
}
if(pHead->val==pHead->next->val){
ListNode* cur_node = pHead->next->next;
int val = pHead->val;
delete pHead;
delete pHead->next;
while(cur_node && cur_node->val==val){
ListNode* be_deleted = cur_node;
cur_node = cur_node->next;
delete be_deleted;
}
return deleteDuplication(cur_node);
}
else{
pHead->next = deleteDuplication(pHead->next);
return pHead;
}
----------
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
class Solution {
public:
ListNode* deleteDuplication(ListNode* pHead)
{
if(pHead==0) return 0;
map<int,int> m;
ListNode* p=pHead;
while(p!=0)
{
m[p->val]++;
p=p->next;
}
//判断head
while(m[pHead->val]>1)
{
pHead=pHead->next;
if(pHead==0) return 0;
}
ListNode* p1=pHead;
ListNode* p2=pHead->next;
while(p2!=0)
{
if(m[p2->val]>1)
{
p2=p2->next;
p1->next=p2;
}
else
{
p2=p2->next;
p1=p1->next;
}
}
return pHead;
*/
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <vector>
#include <string>
#include <queue>
#include <iterator>
#include <stack>
#include <malloc.h>
using namespace std;
#define N 10
typedef struct node
{
int data;
struct node * lchild;
struct node * rchild;
}node;
int count = 0;
#define MIN(a,b) ((a) >= (b) ? (b) : (a))
#define MAX(a,b) ((a) >= (b) ? (a) : (b))
void Init(node *t)
{
t = NULL;
}
node * Insert(node *t , int key)
{
if(t == NULL)
{
node * p;
p = (node *)malloc(sizeof(node));
p->data = key;
p->lchild = NULL;
p->rchild = NULL;
t = p;
}
else
{
if(key < t->data)
t->lchild = Insert(t->lchild, key);
else
t->rchild = Insert(t->rchild, key);
}
return t; //important!
}
node * creat(node *t)
{
int array[N] = {5,7,1,9,4,8,3,6,11,12};
int i;
for (i = 0; i < N; ++i) {
t = Insert(t, array[i]);
}
return t;
}
// / 前序遍历递归
void PreOrder(node * t)
{
if(t != NULL)
{
printf("%d ", t->data);
PreOrder(t->lchild);
PreOrder(t->rchild);
}
}
/*
* 对于任意节点，
* 1.访问节点ｐ，并将节点ｐ入栈
* 2.判断该节点ｐ的左孩子是否为空，若为空，则取栈顶节点并进行出栈操作
* 并将该栈顶节点的右孩子置为当前的ｐ，循环１
若不为空，则将ｐ的左孩子置为当前的节点ｐ
*/
/// 前序遍历非递归
void RePreOrder(node *t)
{
/// / use stack
stack<node *> s;
node *p = t;
while(p != NULL || !s.empty())
{
while( p != NULL )
{
cout << p->data << " ";
s.push(p);
p=p->lchild;
}
if(!s.empty())
{
p = s.top();
s.pop();
p=p->rchild;
}
}
}
void InOrder(node * t) //中序遍历输出
{
if(t != NULL)
{
InOrder(t->lchild);
// printf("%d ", t->data);
printf("%d ", t->data);
InOrder(t->rchild);
}
}
/*
*对于任意节点ｐ
*1.若其左孩子不为空，则将ｐ入栈，并将ｐ的左孩子置为当前的ｐ，
*然后对其节点ｐ再做相同的处理
*2.若左孩子为空，则取栈顶元素进行出栈，访问该节点，然后将当前的ｐ置为该
*栈顶节点的右孩子
*3.直到ｐ为NULL，并且栈为空，则遍历结束
*/
/// 中序遍历非递归
void ReInOrder(node *t)
{
stack<node *> s;
node *p = t;
while(p!= NULL || !s.empty())
{
while(p!=NULL)
{
s.push(p);
p=p->lchild;
}
if(!s.empty())
{
p = s.top();
cout << p->data << " ";
s.pop();
p=p->rchild;
}
}
}
/// 后序遍历
void PostOrder(node * t)
{
if(t != NULL)
{
PostOrder(t->lchild);
PostOrder(t->rchild);
printf("%d ", t->data);
// cout << t->data << " ";
}
}
/*
*
要保证根结点在左孩子和右孩子访问之后才能访问，因此对于任一结点P，先将其入栈。如果P不存在左孩子和右孩子，则可以直接访问它；或者P存在左孩子或者右孩子，但是其左孩子和右孩子都已被访问过了，则同样可以直接访问该结点。若非上述两种情况，则将P的右孩子和左孩子依次入栈，这样就保证了每次取栈顶元素的时候，左孩子在右孩子前面被访问，左孩子和右孩子都在根结点前面被访问。
*/
void RePostOrder(node *t)
{
stack<node *> s;
node *cur;
node *pre = NULL;
s.push(t);
while(!s.empty())
{
cur = s.top();
if((cur->lchild == NULL && cur->rchild==NULL)||
(pre != NULL &&(pre == cur->lchild || pre == cur->rchild))
)
{
cout << cur->data << " ";
s.pop();
pre = cur;
}
else
{
if(cur->rchild != NULL)
s.push(cur->rchild);
if(cur->lchild != NULL)
s.push(cur->lchild);
}
}
}
/// 层次遍历
vector<vector<int> > LevelOrder(node *root)
{
vector<vector<int> > res;
vector<int> v;
if(root == NULL)
return res;
int before = 0;
int curr = 0;
queue<node *> q;
q.push(root);
curr++;
while(!q.empty())
{
before = curr;
curr = 0;
while(before--)
{
node * t = q.front();
if(t->lchild != NULL)
{
q.push(t->lchild);
curr++;
}
if(t->rchild != NULL)
{
q.push(t->rchild);
curr++;
}
v.push_back(t->data);
q.pop();
}
res.push_back(v);
v.clear();
}
return res;
}
node * KthNode(node * root,unsigned int k)
{
if(root)
{
node * ret = KthNode(root->lchild,k);
if(ret)
return ret;
if(++count == k)
return root;
ret = KthNode(root->rchild,k);
if(ret)
return ret;
}
return NULL;
}
void InOrdeTwo(node * t,vector<node *> &v) //中序遍历输出
{
if(t != NULL)
{
InOrdeTwo(t->lchild,v);
v.push_back(t);
InOrdeTwo(t->rchild,v);
}
}
node * KthNodeTwo(node * root,unsigned int k)
{
/// 利用中序遍历，和vector,直接读取倒数第ｋ个就可以了
/// 因为返回来的是节点的指针
if(root == NULL || k <= 0)
return NULL;
vector<node*> vec;
/// 存入节点
InOrdeTwo(root,vec);
if(k > vec.size())
return NULL;
return vec[k-1];
}
int Depth(node *root)
{
return root?(MAX(Depth(root->lchild),Depth(root->rchild)) + 1):0;
}
void FreeNode(node *root)
{
if(root)
{
FreeNode(root->lchild);
FreeNode(root->rchild);
/// 当左右节点都不为空，释放空间
free(root);
}
}
/*
* pre 前序遍历的数组，ｉｎ后遍历 的数组，len 树的长度
递归思想，递归的终止条件是树的长度len == 0
在中序遍历的数组中找到前序数组的第一个字符，记录在中序数组中的位置index.如果找不到，说明前序遍历数组和中序遍历数组有问题
，提示错误信息，退出程序即可;找到index后，新建一个二叉树节点t，t->item= *pre,
然后递归的求t的左孩子和有孩子 递归的左孩子：
void rebuildTree(pre + 1, in, index) 递归的右孩子：
void rebuildTree(pre + (index + 1), in + (index +1), len - (index + 1))
*
*
*
*
*
*/
node *rebuildTree(char *pre,char *in,int len)
{
node *t;
if(len <= 0)
t = NULL;
else
{
int index = 0;
while(index < len && *(pre) != *(in + index))
{
index++;
}
if(index >= len)
{
cout << "error " << endl;
return NULL;
}
t = (node *)malloc(sizeof(node));
t->data = *pre;
t->lchild = rebuildTree(pre+1,in,index);
t->rchild = rebuildTree(pre+(index+1),in +(index+1),len-(index+1));
}
return t;
}
/*
* 根据中序遍历和后序遍历，得到前序遍历
*
根据后序遍历的特点，知道后序遍历最后一个节点为根节点，即为A
观察中序遍历，A左侧CBEDF为A左子树节点，A后侧HGJI为A右子树节点
然后递归的构建A的左子树和后子树
*
*
*
*/
node *rebuildTreeTwo(char *in,char *post,int len)
{
node *t;
if(len <= 0)
t = NULL;
else
{
int index = 0;
while(index < len && *(post + len -1) != *(in + index))
{
index++;
}
// if(index >= len)
// {
// cout << "error " << endl;
// return NULL;
// }
t = (node *)malloc(sizeof(node));
t->data = *(in +index);
t->lchild = rebuildTreeTwo(in,post,index);
t->rchild = rebuildTreeTwo(in+(index+1),post +index,len-(index+1));
}
return t;
}
/**
* @param
* @param 之字形打印二叉树
* @param
* @param
*/
vector<vector<int> > ZhiPrint(node * t)
{
#if 0
vector<vector<int> > result;
stack<node *> stack1,stack2;
if(t!=NULL)
stack1.push(t);
node *p;
while(!stack1.empty() || !stack2.empty()){
vector<int> data;
if(!stack1.empty()){
while(!stack1.empty()){
p = stack1.top();
stack1.pop();
data.push_back(p->data);
if(p->lchild!=NULL)
stack2.push(p->lchild);
if(p->rchild!=NULL)
stack2.push(p->rchild);
}
result.push_back(data);
}
else if(!stack2.empty()){
while(!stack2.empty()){
p = stack2.top();
stack2.pop();
data.push_back(p->data);
if(p->rchild!=NULL)
stack1.push(p->rchild);
if(p->lchild!=NULL)
stack1.push(p->lchild);
}
result.push_back(data);
}
}
return result;
#endif
vector<vector<int> > res;
stack<node *> stack1,stack2;
if(t)
stack1.push(t);
else
return res;
node *p;
/// 一个小小的错误判断是否为空，对于stack2导致自己调试了好久，一定要小心，对于别人的东西，先要理解，然后在跟踪代码
while(!stack1.empty() || !stack2.empty()){
///对于临时的数据data,每次的使用都会清空，所以我们在两个栈交换的同时，转到while在重新定义变量
vector<int> data;
if(!stack1.empty()){
while(!stack1.empty()){
p = stack1.top();
stack1.pop();
data.push_back(p->data);
if(p->lchild!=NULL)
stack2.push(p->lchild);
if(p->rchild!=NULL)
stack2.push(p->rchild);
}
res.push_back(data);
}
else if(!stack2.empty()){
while(!stack2.empty()){
p = stack2.top();
stack2.pop();
data.push_back(p->data);
if(p->rchild!=NULL)
stack1.push(p->rchild);
if(p->lchild!=NULL)
stack1.push(p->lchild);
}
res.push_back(data);
}
}
return res;
}
/*
* 代码的规范
* while(exper){
* ......;
}
*/
int SumNode(node *root)
{
int count = 0;
if(root == NULL)
return 0;
else if(root->rchild == NULL && root->lchild == NULL)
return 1;
else
count = SumNode(root->lchild)+SumNode(root->rchild);
return count;
}
int main()
{
node * t = NULL;
t = creat(t);
cout << "preorder " << endl;
/// PreOrder
cout << "--------------" << endl;
PreOrder(t);
cout << endl;
RePreOrder(t);
cout << endl;
cout << "--------------" << endl;
cout << endl;
cout << "inorder" << endl;
cout << "--------------" << endl;
InOrder(t);
cout << endl;
ReInOrder(t);
cout << endl;
cout << "--------------" << endl;
cout << "---post-----------" << endl;
PostOrder(t);
cout << endl;
cout << "---repost-----------" << endl;
RePostOrder(t);
cout << endl;
#if 0
cout << endl;
cout << endl;
cout << "postorder" << endl;
cout << "--------------" << endl;
PostOrder(t);
cout << endl;
cout << endl;
cout << "--------------" << endl;
#endif
/// 之子打印二叉树
vector<vector<int> > zhi;
zhi = ZhiPrint(t);
cout << "liang***" << endl;
for (unsigned int i = 0; i < zhi.size(); ++i) {
for (unsigned int j = 0; j < zhi[i].size(); ++j) {
cout << zhi[i][j] << " ";
}
cout << endl;
}
cout << "liang***" << endl;
cout << "sum node:" << endl;
int sum = SumNode(t);
cout << sum << endl;
// SumNode(t,sum);
#if 0
vector<vector<int> > layer = LevelOrder(t);
for (int i = 0; i < layer.size(); ++i) {
for (int j = 0; j < layer[i].size(); ++j){
cout << layer[i][j] << " ";
}
cout << endl;
}
for (int i = 0; i < layer.size(); ++i) {
if(i % 2 == 0)
{
for (int j = 0; j < layer[i].size(); ++j)
cout << layer[i][j] << " ";
}
else
{
for (int k = layer[i].size()-1; k >=0 ; k--)
cout << layer[i][k] << " ";
}
cout << endl;
}
cout << "2 node :";
node * k = KthNode(t,2);
cout << k->data << endl;
cout << "3 node :";
node * k1 = KthNodeTwo(t,3);
cout << k1->data << endl;
cout << "the tree depth : " << Depth(t) << endl;
/// 根据前序和中序遍历得到二叉树
char pre[] = "abcdefgh";
char in[] = "cbedfagh";
node *bt = NULL;
bt = rebuildTree(pre,in,8);
InOrder(bt);
/// 根据后序和中序遍历得到二叉树
char str_inder[] = "cbedfahgji";
char str_post[] = "cefdbhjiga";
node *bt1 = NULL;
bt1 = rebuildTreeTwo(str_inder,str_post,10);
cout << endl;
InOrder(bt1);
FreeNode(bt);
bt=NULL;
FreeNode(bt1);
bt1=NULL;
#endif
FreeNode(t);
t = NULL;
return 0;
}
ipyt to pdf
######################################### # File name :npynb2pdf.sh # Author :kangkangliang # File desc : # Mail :liangkangkang@paag.com # Create time :2016-03-08 ######################################### #!/usr/bin/bash ipython nbconvert --to latex --post PDF 00-classification.ipynb ipython nbconver --to PDF 00-classification.ipynb ipython nbconvert --to latex Example.ipynb ipython nbconvert --to latex 00-classification.ipynb xelatex 00-classification.tex
浙公网安备 33010602011771号