data structure test
1、设计算法,对带头结点的单链表实现就地逆置。并给出单链表的存储结构(数据类型)的定义。
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <ctime>
using namespace std; typedef char ElemType; typedef struct Node{
ElemType data;
struct Node *next;
}Node, *LinkList; LinkList CreateList()
{
LinkList L;
ElemType c;
L = (LinkList)malloc(sizeof(Node));
L->next = NULL;
Node *p , *tail;
tail = L;
c = getchar();
while(c != '#')
{
p = (Node *)malloc(sizeof(Node));
p->data = c;
tail->next = p;
tail = p;
c = getchar();
}
tail->next = NULL;
return L;
} void ShowList(LinkList L)
{
Node *p;
p = L->next;
while(p != NULL)
{
cout << p->data << " ";
p = p->next;
}
cout << endl;
} void ReverseList(LinkList L)
{
Node *p, *q;
p = L->next;
L->next = NULL;
while(p != NULL)
{
q = p->next;
p->next = L->next;
L->next = p;
p = q;
} } int main()
{
LinkList L;
L = CreateList();
ShowList(L); ReverseList(L);
ShowList(L);
return 0;
}
2、编写递归算法,将二叉树中所有结点的左、右子树相互交换。并给出算法中使用的二叉树的存储结构(数据类型)的定义。
typedef struct BiNode
{
char data;
struct BiNode *left;
struct BiNode *right;
}BiNode, *BiTree;
BiNode* Exchange(BiNode* T)
{
BiNode* p;
if(NULL==T || (NULL==T->lchild && NULL==T->rchild))
return T;
p = T->lchild;
T->lchild = T->rchild;
T->rchild = p;
if(T->lchild)
{
T->lchild = Exchange(T->lchild);
}
if(T->rchild)
{
T->rchild = Exchange(T->rchild);
}
return T;
}
3、折半查找算法。
#include<iostream>
#include<cstdio>
using namespace std;
int search(int *array, int n, int target)
{
int low = 0;
int high = n - 1;
if( low > high ) return -1;
while( low <= high )
{
int mid = low + ( high - low ) / 2;
if( array[mid] < target ) low = mid + 1;
else if( array[mid] > target ) high = mid - 1;
else return mid;
}
}
int main()
{
int a[] = {1,2,3,4,5,6,7,8,9};
cout<<search(a,9,4)<<endl;
return 0;
}
4、数据结构课本P74,习题2、13
void DeleteNode( ListNode *s)
{//删除单循环链表中指定结点的直接前趋结点
ListNode *p, *q;
p=s;
while( p->next->next!=s)
p=p->next;
//删除结点
q=p->next;
p->next=q->next;
free(p); //释放空间
}
5、统计叶子结点数。P168
#include<iostream>
#include<queue>
#include <cstdlib>
#include <cstdio>
using namespace std;
typedef struct BiNode
{
char data;
struct BiNode *left;
struct BiNode *right;
}BiNode, *BiTree;
int sum = 0;
void CreateBinaryTree(BiTree &T)//二叉树建立 abc,,de,g,,f,,,
{
//T = (BiNode*) malloc (sizeof(BiNode));
T = new BiNode;
cin >> T->data;
if(T->data == ',')
{
T = NULL;
}
if(T != NULL)
{
CreateBinaryTree(T->left);
CreateBinaryTree(T->right);
}
}
void PreOrder(BiTree T)//前序遍历
{
if(T != NULL)
{
cout << T->data;
PreOrder(T->left);
PreOrder(T->right);
}
}
void InOrder(BiTree T)//中序遍历
{
if(T != NULL)
{
InOrder(T->left);
cout << T->data;
InOrder(T->right);
}
}
void PostOrder(BiTree T)//后序遍历
{
if(T != NULL)
{
PostOrder(T->left);
PostOrder(T->right);
cout << T->data;
}
}
void LevOrder(BiTree T)//层次遍历
{
if(T != NULL)
{
BiTree p = T;
queue<BiTree>que;
que.push(p);
while(!que.empty())
{
p = que.front();
cout << p->data;
que.pop();
if(p->left != NULL)
{
que.push(p->left);
}
if(p->right != NULL)
{
que.push(p->right);
}
}
}
}
int Size(BiTree T)//计算二叉树节点数
{
if(T != NULL)
{
if(T->left == NULL && T->right == NULL)
{
sum++;
}
Size(T->left);
Size(T->right);
}
return sum;
}
int Deep(BiTree T)//计算二叉树深度
{
int m, n;
if(T == NULL) return 0;
m = Deep(T->left);
n = Deep(T->right);
if(m > n) return m + 1;
else return n + 1;
}
int main(void)
{
BiTree T;
CreateBinaryTree(T); cout << "前序遍历结果为:" << endl;
PreOrder(T);
cout << endl << endl; cout << "中序遍历结果为:" << endl;
InOrder(T);
cout << endl << endl; cout << "后序遍历结果为:" << endl;
PostOrder(T);
cout << endl << endl; cout<<"层次遍历结果为:"<<endl;
LevOrder(T);
cout << endl << endl; cout << "二叉树叶节点个数为:" << Size(T)<<endl;
cout << "二叉树深度数为:" << Deep(T) << endl;
system("pause");
return 0;
}
6、顺序表的合并。
#define MAXSIZE 100
typedef int ElemType; typedef struct SeqList
{
ElemType elem[MAXSIZE];
int last;
}SeqList;
void mergeList(SeqList *LA, SeqList * LB, SeqList *LC)
{
int i, j, k;
i = j = k = 0;
while (i <= LA->last && j <= LB->last)
{
if (LA->elem[i] <= LB->elem[j])
{
LC->elem[k++] = LA->elem[i++];
}
else
{
LC->elem[k++] = LB->elem[i++];
}
}
while (i <= LA->last)
{
LC->elem[k++] = LA->elem[i++];
}
while (j <= LB->last)
{
LC->elem[k++] = LB->elem[j++];
}
LC->last = LA->last + LB->last + 1;
}
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