第一百一十二天学习记录：数据结构与算法基础：循环链表和双向链表以及线性表应用（王卓教学视频）

循环链表

带尾指针循环链表的合并

双向链表

单链表、循环链表和双向链表的时间效率比较

顺序表和链表的比较

链式存储结构的优点

1、结点空间可以动态申请和释放；
2、数据元素的逻辑次序靠结点的指针来指示，插入和删除时不需要移动数据元素。

链式存储结构的缺点

1、存储密度小，每个结点的指针域需额外占用存储空间。当每个结点的数据域所占字节不多时，指针域所占存储空间的比重显得很大。
2、链式存储结构是非随机存取结构。对任一节点的操作都要从头指针依指针链查找到该结点，这增加了算法的复杂度。

线性表应用

用C语言实现：

#include 

void mergeArrays(int arr1[], int size1, int arr2[], int size2, int mergedArray[]) {
    int i, j, k, duplicate;

    // 将arr1的元素插入合并数组
    for (i = 0; i < size1; i++) {
        mergedArray[i] = arr1[i];
    }
    k = size1;

    // 将arr2中非重复元素插入合并数组
    for (i = 0; i < size2; i++) {
        duplicate = 0;
        for (j = 0; j < size1; j++) {
            if (arr2[i] == arr1[j]) {
                duplicate = 1;
                break;
            }
        }
        if (!duplicate) {
            mergedArray[k++] = arr2[i];
        }
    }
}

int main() {
    int arr1[] = {1, 2, 3, 4, 5};
    int size1 = sizeof(arr1) / sizeof(arr1[0]);

    int arr2[] = {4, 5, 6, 7, 8};
    int size2 = sizeof(arr2) / sizeof(arr2[0]);

    // 计算合并数组的最大长度
    int maxSize = size1 + size2;

    int mergedArray[maxSize];

    mergeArrays(arr1, size1, arr2, size2, mergedArray);

    // 输出合并并去重后的数组
    for (int i = 0; i < size1 + size2; i++) {
        printf("%d ", mergedArray[i]);
    }
    printf("\n");

    return 0;
}

实现代码如下：

#include
using namespace std;

struct Mylist
{
	int m_int;
	Mylist* myl;
};

int main()
{
	Mylist* la = new Mylist;
	la->myl = new Mylist;
	la->myl->myl = new Mylist;
	la->myl->myl->myl = new Mylist;
	la->myl->myl->myl->myl = NULL;
	la->myl->m_int = 1;
	la->myl->myl->m_int = 7;
	la->myl->myl->myl->m_int = 8;
	
	Mylist* lb = new Mylist;
	lb->myl = new Mylist;
	lb->myl->myl = new Mylist;
	lb->myl->myl->myl = new Mylist;
	lb->myl->myl->myl->myl = new Mylist;
	lb->myl->myl->myl->myl->myl = new Mylist;
	lb->myl->myl->myl->myl->myl->myl = new Mylist;
	lb->myl->myl->myl->myl->myl->myl->myl = NULL;
	lb->myl->m_int = 2;
	lb->myl->myl->m_int = 4;
	lb->myl->myl->myl->m_int = 6;
	lb->myl->myl->myl->myl->m_int = 8;
	lb->myl->myl->myl->myl->myl->m_int = 10;
	lb->myl->myl->myl->myl->myl->myl->m_int = 11;

	Mylist* pc = la;
	Mylist* lc = la;
	Mylist* pa = la->myl;
	Mylist* pb = lb->myl;

	while (pa&&pb)
	{
		if (pa->m_int <= pb->m_int)
		{
			pc->myl = pa; pc = pa; pa = pa->myl;
		}
		else
		{
			pc->myl = pb; pc = pb; pb = pb->myl;
		}
	}
	pc->myl = pa ? pa : pb;
	delete lb;
	pc = lc;
	while (pc->myl)
	{
		cout << pc->myl->m_int << " ";
		pc = pc->myl;
	}
	cout << endl;
	return 0;
}

案例分析与实现

实现代码：

本来以为很简单，全部写在main函数里，结果……
本代码笔者尽可能将会导致程序出错以及内存泄漏的情况规避了。如果还有有缺陷的地方欢迎大佬们指出。

#include
using namespace std;

struct Poly
{
	int c;
	int e;
	Poly* pnext;
};

int main()
{
	cout << "本程序用链表方式实现两个一元多项式相加" << endl;
	cout << "由低阶到高阶依次输入一元多项式的系数和指数" << endl;
	cout << "用空格分隔，回车结束：(系数为0不需要输入)" << endl;
	cout << "比如:y(x)=3x^0+4x^1+x^3  输入：3 0 4 1 1 3" << endl;
	cout << "提示：输入任意非数字键退出" << endl;
	int input;
	int ctmp = 0;
	int etmp = 0;
	int tmp = 0;
	int pta = 0;
	int ptb = 0;
	Poly* Pahead = new Poly;
	Pahead->pnext = NULL;
	Poly* Pa = Pahead;
	Poly* Pbhead = new Poly;
	Pbhead->pnext = NULL;
	Poly* Pb = Pahead;
	Poly* temp = NULL;
	Poly* Pc = NULL;
	while (true)
	{
		Pa = Pahead;
		Pa = Pa->pnext;
		while (Pa)
		{
			//cout << "Pa删除一项" << endl;
			temp = Pa->pnext;
			delete Pa;
			Pa = temp;
		}
		Pa = Pahead;
		Pb = Pbhead;
		Pb = Pb->pnext;
		Pahead->pnext = NULL;
		while (Pb)
		{
			//cout << "Pb删除一项" << endl;
			temp = Pb->pnext;
			delete Pb;
			Pb = temp;
		}
		Pb = Pbhead;
		Pbhead->pnext = NULL;
		cout << "下面请输入第一个一元多项式一共有几项" << endl;
		cout << "=>";
		while (true)
		{
			cin >> input;
			if (cin.fail())
			{
				cin.clear();
				cout << "已退出，欢迎再次使用。" << endl;
				Pa = Pahead;
				Pb = Pbhead;
				if (Pahead != NULL)
				{
					//cout << "删除Pahead。" << endl;
					Pa = Pahead;
					while (Pa)
					{
						temp = Pa->pnext;
						delete Pa;
						Pa = temp;
					}
				}
				if (Pbhead != NULL)
				{
					//cout << "删除Pbhead。" << endl;
					Pb = Pbhead;
					while (Pb)
					{
						temp = Pb->pnext;
						delete Pb;
						Pb = temp;
					}
				}
				return 0;
			}
			if (input <= 0)
			{
				cout << "请至少输入一项" << endl;
				continue;
			}
			if (input > 100)
			{
				cout << "最大不要超出100项……这么多的多项式爱因斯坦见了都得捂脸……" << endl;
				continue;
			}
			pta = input * 2;
			break;
		}
		cout << "下面请输入第一个一元多项式" << endl;
		cout << "=>";
		while (pta--)
		{
			cin >> input;
			if (cin.fail())
			{
				cin.clear();
				cout << "已退出，欢迎再次使用。" << endl;
				Pa = Pahead;
				Pb = Pbhead;
				if (Pahead != NULL)
				{
					//cout << "删除Pahead。" << endl;
					Pa = Pahead;
					while (Pa)
					{
						temp = Pa->pnext;
						delete Pa;
						Pa = temp;
					}
				}
				if (Pbhead != NULL)
				{
					//cout << "删除Pbhead。" << endl;
					Pb = Pbhead;
					while (Pb)
					{
						temp = Pb->pnext;
						delete Pb;
						Pb = temp;
					}
				}
				return 0;
			}
			if (tmp == 0)
			{
				ctmp = input;
				++tmp;
				continue;
			}
			else
			{
				etmp = input;
				Pa->pnext = new Poly;
				Pa->pnext->c = ctmp;
				Pa->pnext->e = etmp;
				Pa = Pa->pnext;
				Pa->pnext = NULL;
				--tmp;
			}
		}
		int availableChars = (int)cin.rdbuf()->in_avail();
		if (availableChars)
		{
			cin.ignore(numeric_limits<streamsize>::max(), '\n');
		}
		cout << "第一个一元多项式:" << endl;
		cout << "y(x)=";
		Pa = Pahead;
		cout << Pa->pnext->c << "x^" << Pa->pnext->e;
		Pa = Pa->pnext;
		while (Pa->pnext)
		{
			cout << "+";
			cout << Pa->pnext->c << "x^" << Pa->pnext->e;
			Pa = Pa->pnext;
		}
		cout << endl;
		//第二个
		Pb = Pbhead;
		cout << "下面请输入第二个一元多项式一共有几项" << endl;
		cout << "=>";
		while (true)
		{
			cin >> input;
			if (cin.fail())
			{
				cin.clear();
				cout << "已退出，欢迎再次使用。" << endl;
				Pa = Pahead;
				Pb = Pbhead;
				if (Pahead != NULL)
				{
					//cout << "删除Pahead。" << endl;
					Pa = Pahead;
					while (Pa)
					{
						temp = Pa->pnext;
						delete Pa;
						Pa = temp;
					}
				}
				if (Pbhead != NULL)
				{
					//cout << "删除Pbhead。" << endl;
					Pb = Pbhead;
					while (Pb)
					{
						temp = Pb->pnext;
						delete Pb;
						Pb = temp;
					}
				}
				return 0;
			}
			if (input <= 0)
			{
				cout << "请至少输入一项" << endl;
				continue;
			}
			if (input > 100)
			{
				cout << "最大不要超出100项……这么多的多项式爱因斯坦见了都得捂脸……" << endl;
				continue;
			}
			ptb = input * 2;
			break;
		}
		cout << "下面请输入第二个一元多项式" << endl;
		cout << "=>";
		while (ptb--)
		{
			cin >> input;
			if (cin.fail())
			{
				cin.clear();
				cout << "已退出，欢迎再次使用。" << endl;
				Pa = Pahead;
				Pb = Pbhead;
				if (Pahead != NULL)
				{
					//cout << "删除Pahead。" << endl;
					Pa = Pahead;
					while (Pa)
					{
						temp = Pa->pnext;
						delete Pa;
						Pa = temp;
					}
				}
				if (Pbhead != NULL)
				{
					//cout << "删除Pbhead。" << endl;
					Pb = Pbhead;
					while (Pb)
					{
						temp = Pb->pnext;
						delete Pb;
						Pb = temp;
					}
				}
				return 0;
			}
			if (tmp == 0)
			{
				ctmp = input;
				++tmp;
				continue;
			}
			else
			{
				etmp = input;
				Pb->pnext = new Poly;
				Pb->pnext->c = ctmp;
				Pb->pnext->e = etmp;
				Pb = Pb->pnext;
				Pb->pnext = NULL;
				--tmp;
			}
		}
		availableChars = (int)cin.rdbuf()->in_avail();
		if (availableChars)
		{
			cin.ignore(numeric_limits<streamsize>::max(), '\n');
		}
		cout << "第二个一元多项式:" << endl;
		cout << "z(x)=";
		Pb = Pbhead;
		cout << Pb->pnext->c << "x^" << Pb->pnext->e;
		Pb = Pb->pnext;
		while (Pb->pnext)
		{
			cout << "+";
			cout << Pb->pnext->c << "x^" << Pb->pnext->e;
			Pb = Pb->pnext;
		}
		cout << endl;
		//求和
		Pa = Pahead->pnext;
		Pc = Pahead;
		Pb = Pbhead->pnext;
		while (Pa&&Pb)
		{
			if (Pa->e < Pb->e)
			{
				Pc->pnext = Pa; Pc = Pa; Pa = Pa->pnext;
			}
			else if (Pa->e > Pb->e)
			{
				Pc->pnext = Pb; Pc = Pb; Pb = Pb->pnext;
			}
			else
			{
				if ((Pa->c + Pb->c) == 0)
				{
					temp = Pb;
					Pb = Pb->pnext;
					delete temp;
					temp = Pa;
					Pa = Pa->pnext;
					delete temp;
				}
				else
				{
					Pc->pnext = Pa; Pc = Pa; Pa = Pa->pnext;
					Pc->c += Pb->c;
					temp = Pb;
					Pb = Pb->pnext;
					delete temp;
				}
			}
		}
		Pc->pnext = Pa ? Pa : Pb;
		Pbhead->pnext = NULL;//必须设置为NULL，要不然下次对Pbhead进行释放时会造成野指针
		cout << "二个一元多项式之和为:" << endl;
		Pc = Pahead;
		cout << "y(x)+z(x)=";
		if (!Pc->pnext)//存在两个多项式相加为零的情况！
		{
			cout << 0 << endl;
			Pc = Pahead;
			continue;
		}
		cout << Pc->pnext->c << "x^" << Pc->pnext->e;
		Pc = Pc->pnext;
		while (Pc->pnext)
		{
			cout << "+";
			cout << Pc->pnext->c << "x^" << Pc->pnext->e;
			Pc = Pc->pnext;
		}
		cout << endl;
		Pc = Pahead;
	}
	return 0;
}

输出结果：

案例实现方式与第一百零六天写的资源管理系统类似，这里不再赘述。