基算要素

coding

tip

input&output

stdin stdout
scanf() printf()
gets() puts()
getchar() putchar()

让gcc的cin和scanf一样快

ios::sync_with_stdio(false);
std::cin.tie(0);

不要用局部变量地址返回地址

int* func(){
    int arr[2] = {}
    return arr;
}

这是错误的，从函数本身的意义来说，该函数应该返回一个指针，但是函数体内的arr是一个临时变量，所以返回arr后会由于临时变量的销毁导致函数返回的地址是一个无意义地址。所以要将int arr静态化，或者改为全局函数

int arr1[2] = {};
int* func1(){

    arr1[0] = 0;
    arr1[1] = 0;
    return arr1;
}


int* func2(){
    static int arr2[2] = {};
    return arr2;
}

一个简单的函数辅助理解c++中指针和数组的联系

#include <iostream>
#include <stdio.h>
using namespace std;
int t = 0;
void ceshi(char *p)
{
    t++;
    if (t > 11)
    {
        return;
    }
    else
    {
        cout << p[0];
        cout << endl;
        ceshi(p + 1);
    }
}
int main()
{
    
    char str[] = "0123456789";
    ceshi(str);
    return 0;
}

algorithm

指针&单链表链表 

sudtoj-1138:单链表操作A

Description 输入n个整数，先按照数据输入的顺序建立一个带头结点的单链表，再输入一个数据m,将单链表中的值为m的结点全部删除。分别输出建立的初始单链表和完成删除后的单链表。

Input
第一行输入数据个数n；
第二行依次输入n个整数；
第三行输入欲删除数据m。
Output
第一行输出原始单链表的长度；
第二行依次输出原始单链表的数据；
第三行输出完成删除后的单链表长度；
第四行依次输出完成删除后的单链表数据。
Sample
Input

10
56 25 12 33 66 54 7 12 33 12
12

Output

10
56 25 12 33 66 54 7 12 33 12
7
56 25 33 66 54 7 33

solutions

#include <iostream>
#include <stdio.h>
using namespace std;

bool equal(int a, int b)
{
    return a == b;
}

struct node
{
    int data;
    node *next = NULL;
};

node *create(int i)
{
    node *head;
    node *p;
    head = new node();
    p = head;
    while (i--)
    {
        p->next = new node();
        p = p->next;
        cin >> p->data;
    }
    return head;
}

/**
 * 创建逆序链表
 */
node *createR(int i)
{

    node *head = new node();
    node *p;
    while (i--)
    {
        p = new node();
        cin >> p->data;
        p->next = head->next;
        head->next = p;
    }
    return head;
}

node *delRepeatNode(node *head)
{
    node *p, *q, *t4del;
    p = head;
    while (p->next != NULL)
    {
        p = p->next;
        q = p;
        while (q->next != NULL)
        {
            if (equal(q->next->data, p->data))
            {
                t4del = q->next;
                q->next = q->next->next;
                delete (t4del);
            }
            else
            {
                q = q->next;
            }
        }
    }
    return head;
}

node *delByData(node *head, int delData)
{
    node *p, *t4del;
    p = head;
    while (p->next != NULL)
    {
        if (p->next->data == delData)
        {

            t4del = p->next;
            p->next = p->next->next;
            delete (t4del);
        }
        else
        {
            p = p->next;
        }
    }
    return head;
}
int getCount(node *head)
{
    node *p = head;
    int counter = 0;
    while (p->next != NULL)
    {
        p = p->next;
        counter++;
    }
    return counter;
}
void printNodeList(node *head)
{
    node *p = head;
    int counter = 0;
    while (p->next != NULL)
    {
        cout << p->next->data;
        if (p->next->next != NULL)
        {
            cout << " ";
        }
        p = p->next;
    }
    cout << endl;
}
void testA()
{
    int n = 0;
    int waitDelData = 0;
    cin >> n;
    node *head = create(n);
    cin >> waitDelData;
    cout << getCount(head) << endl;
    printNodeList(head);
    head = delByData(head, waitDelData);
    cout << getCount(head) << endl;
    printNodeList(head);
}
void testB()
{
    int n = 0;
    cin >> n;
    node *head = createR(n);
    cout << getCount(head) << endl;
    printNodeList(head);
    head = delRepeatNode(head);
    cout << getCount(head) << endl;
    printNodeList(head);
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    testA();
    // testB();
    return 0;
}

sdutoj-1139:单链表操作B

Description
按照数据输入的相反顺序（逆位序）建立一个单链表，并将单链表中重复的元素删除（值相同的元素只保留最后输入的一个）。
Input
第一行输入元素个数n；
第二行输入n个整数。
Output
第一行输出初始链表元素个数；
第二行输出按照逆位序所建立的初始链表；
第三行输出删除重复元素后的单链表元素个数；
第四行输出删除重复元素后的单链表。
Sample
Input

10
21 30 14 55 32 63 11 30 55 30

Output

10
30 55 30 11 63 32 55 14 30 21
7
30 55 11 63 32 14 21

解答为上述代码的testB函数

sdutoj-2054:双向链表

#include <iostream>
#include <stdio.h>
using namespace std;

bool equal(int a, int b)
{
    return a == b;
}

struct node
{
    int data;
    node *next = NULL;
    node *last = NULL;
};

node *create(int i)
{
    node *head;
    node *p;
    head = new node();
    p = head;
    while (i--)
    {
        p->next = new node();
        p->next->last = p;
        p = p->next;
        cin >> p->data;
    }
    return head;
}

void findLastAndNext(node *head, int data)
{
    node *p = head;
    while (p->next != NULL)
    {
        if (p->next->data == data)
        {
            if (p != head)
            {
                cout << p->data;
            }
            if (p != head && p->next->next != NULL)
            {
                cout << " ";
            }
            if (p->next->next != NULL)
            {
                cout << p->next->next->data;
            }
            cout<<endl;
        }
        p = p->next;
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n = 0;
    int m = 0;
    cin >> n;
    cin >> m;
    node *head = create(n);
    int findData;
    while (m--)
    {
        cin >> findData;
        findLastAndNext(head, findData);
    }
    return 0;
}

sdutoj-2121:链表排序

#include <iostream>
using namespace std;

struct node
{
    int data;
    node *next;
};
node *creat(int n)
{
    node *head, *p, *tail;
    head = new node;
    head->next = NULL;
    tail = head;
    while (n--)
    {
        p = new node;
        cin >> p->data;
        p->next = NULL;
        tail->next = p;
        tail = p;
    }
    return (head);
}

node *sortlist(node *head)
{
    node *q, *p;
    int temp;
    for (p = head->next; p != NULL; p = p->next)
    {
        for (q = p->next; q != NULL; q = q->next)
        {
            if (p->data > q->data)
            {
                temp = q->data;
                q->data = p->data;
                p->data = temp;
            }
        }
    }
    return head;
}
void printNodeList(node *head)
{
    node *r;
    r = head;
    while (r->next->next != NULL)
    {
        cout << r->next->data << " ";
        r = r->next;
    }
    cout << r->next->data << endl;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    node *head, *r;
    int n;
    cin >> n;
    head = creat(n);
    head = sortlist(head);
    printNodeList(head);
    return 0;
}

循环链表（略）

贪心

活动选择问题

Description
sdut 大学生艺术中心每天都有n个活动申请举办，但是为了举办更多的活动，必须要放弃一些活动，求出每天最多能举办多少活动。

Input
输入第一行为申请的活动数n(n<100)，从第2行到n+1行，每行两个数，是每个活动的开始时间b，结束时间e；

Output
输出每天最多能举办的活动数。

Sample
Input 

12
15 20
15 19
8 18
10 15
4 14
6 12
5 10
2 9
3 8
0 7
3 4
1 3

Output 

5

#include <iostream>
#include <stdio.h>
using namespace std;

struct actionTime
{
    int startTime = 0;
    int endTime = 0;
};

int main()
{
    int n = 0;
    actionTime actionTimeList[200] = {};
    actionTime tmpActionTime;
    while (int scanfResult = scanf("%d", &n))
    {
        if (scanfResult == EOF)
        {
            break;
        }
        for (int i = 0; i < n; i++)
        {
            cin >> actionTimeList[i].startTime >> actionTimeList[i].endTime;
        }
        for (int i = 0; i < n - 1; i++)
        {
            for (int j = 0; j < n - i - 1; j++)
            {
                if (actionTimeList[j].endTime > actionTimeList[j + 1].endTime)
                {
                    tmpActionTime = actionTimeList[j];
                    actionTimeList[j] = actionTimeList[j + 1];
                    actionTimeList[j + 1] = tmpActionTime;
                }
            }
        }
        // cout << "##############" << endl;
        // for (int i = 0; i < n; i++)
        // {
        //     cout << actionTimeList[i].startTime << " " << actionTimeList[i].endTime << endl;
        // }

        int j = 0;
        int sum = 1;
        for (int i = 1; i < n; i++)
        {
            if (actionTimeList[i].startTime >= actionTimeList[j].endTime)
            {
                sum++;
                j = i;
            }
        }
        cout << sum << endl;
    }
    return 0;
}

[]完全背包

[]动态规划(01背包)

二叉树 

sdutoj-2136:数据结构实验之二叉树的建立与遍历

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

struct node
{
    char data;
    node *left = NULL;
    node *right = NULL;
};

char charList[60] = "";
int charListP = 0;
node *create()
{

    char data = charList[charListP];
    charListP++;
    if (data != ',')
    {
        node *p = new node();
        p->data = data;
        p->left = create();
        p->right = create();
        return p;
    }
    else
    {
        return NULL;
    }
}

void proOrder(node *p)
{
    cout << p->data;
    if (p->left)
    {
        proOrder(p->left);
    }
    if (p->right)
    {
        proOrder(p->right);
    }
}

void inOrder(node *p)
{

    if (p->left)
    {
        inOrder(p->left);
    }

    cout << p->data;
    if (p->right)
    {
        inOrder(p->right);
    }
}
void lastOrder(node *p)
{
    if (p->left)
    {
        lastOrder(p->left);
    }
    if (p->right)
    {
        lastOrder(p->right);
    }
    cout << p->data;
}
int leafCounter = 0;
void initLeafCounter()
{
    leafCounter = 0;
}
void getLeafCount(node *p)
{
    if (p->left)
    {
        getLeafCount(p->left);
    }
    if (p->right)
    {
        getLeafCount(p->right);
    }
    if (p->left == NULL && p->right == NULL)
    {
        leafCounter++;
    }
}

int deepCounter = 0;
void initDeepCounter()
{
    deepCounter = 0;
}
void getDeep(node *p, int nowDeep = 0)
{
    nowDeep += 1;
    if (p->left)
    {
        getDeep(p->left, nowDeep);
    }
    if (p->right)
    {
        getDeep(p->right, nowDeep);
    }
    if (p->left == NULL && p->right == NULL)
    {
        if (nowDeep > deepCounter)
        {
            deepCounter = nowDeep;
        }
    }
}
int main()
{
    while (cin >> charList)
    {
        charListP = 0;
        node *head;
        head = create();
        // proOrder(head);
        // cout << endl;
        inOrder(head);
        cout << endl;
        lastOrder(head);
        cout << endl;
        initLeafCounter();
        getLeafCount(head);
        cout << leafCounter << endl;
        initDeepCounter();
        getDeep(head);
        cout << deepCounter << endl;
    }
}

数据结构实验之求二叉树后序遍历和层次遍历

其中层次遍历有循环多次比较深度和队列两种方式实现

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

struct node
{
    char data;
    node *left = NULL;
    node *right = NULL;
};

/**
 * 一切以构造先序构造条件为优先
 */
node *create(char *pre, char *in, int len)
{
    bool isNode = len > 0;
    // 判断当前pre字符串是否为空
    if (isNode)
    {
        /***
         * ROOT LEFT***** RIGHT*****
         * LEFT***** ROOT RIGHT*****
         */
        node *p = new node();
        p->data = pre[0];
        int leftNodeNum = 0;
        int rightNodeNum = 0;

        while (pre[0] != in[leftNodeNum])
        {
            leftNodeNum++;
        }
        // 下标0为起点所以找到root时正好为left的长度

        rightNodeNum = len - leftNodeNum - 1;

        p->left = create(pre + 1, in, leftNodeNum);

        p->right = create(pre + 1 + leftNodeNum, in + leftNodeNum + 1, rightNodeNum);

        return p;
    }
    else
    {
        return NULL;
    }
}

void lastOrder(node *head)
{
    if (head != NULL)
    {
        lastOrder(head->left);
        lastOrder(head->right);
        cout << head->data;
    }
}
int deepCounter = 0;
void getDeep(node *p, int nowDeep = 0)
{
    nowDeep += 1;
    if (p->left)
    {
        getDeep(p->left, nowDeep);
    }
    if (p->right)
    {
        getDeep(p->right, nowDeep);
    }
    if (p->left == NULL && p->right == NULL)
    {
        if (nowDeep > deepCounter)
        {
            deepCounter = nowDeep;
        }
    }
}

int findDeep = 0;
void deepOrder(node *head, int nowDeep)
{
    if (nowDeep > findDeep)
    {
        return;
    }
    if (head != NULL)
    {
        if (nowDeep == findDeep)
        {
            cout << head->data;
        }
        deepOrder(head->left, nowDeep + 1);
        deepOrder(head->right, nowDeep + 1);
    }
}

void initDeepFunction()
{
    deepCounter = 0;
    findDeep = 0;
}

struct queueList
{
    node *data;
    queueList *next;
};

queueList *tail;
queueList *queueHead;
void insertQueueList(node *data)
{
    tail->data = data;
    tail->next = new queueList();
    tail = tail->next;
}

node *pushQueueList()
{
    queueList *waitDelNode = queueHead->next;
    node *data = waitDelNode->data;
    queueHead->next = queueHead->next->next;
    delete (waitDelNode);
    return data;
}

void deepOrderByQueue(node *treeHead)
{
    queueHead = new queueList();
    queueHead->next = new queueList();
    tail = queueHead->next;
    insertQueueList(treeHead);
    while (queueHead->next != tail)
    {

        node *waitShow = pushQueueList();
        if (waitShow->left != NULL)
        {
            insertQueueList(waitShow->left);
        }
        if (waitShow->right != NULL)
        {
            insertQueueList(waitShow->right);
        }
        cout << waitShow->data;
    }
}

int main()
{

    char proStr[100];
    char inStr[100];
    int num = 0;
    cin >> num;
    while (num--)
    {
        cin >> proStr;
        cin >> inStr;
        node *head;
        head = create(proStr, inStr, strlen(proStr));
        lastOrder(head);
        cout << endl;

        // 利用层级多次调用遍历函数
        // initDeepFunction();
        // getDeep(head);
        // for (findDeep = 1; findDeep <= deepCounter; findDeep++)
        // {
            // deepOrder(head, 1);
        // }
        // cout << endl;

        // 利用队列深度遍历
        deepOrderByQueue(head);
        cout << endl;
    }
    return 0;
}

多叉树 

多叉树构建英文字典

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