数据结构、算法、程序：已知一个带头结点的单链表La，其中存放着一组非零整数。 C++数据结构题：一个带头节点的单链表乘以一组整形数，分裂链...
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#include
#include
typedef struct node
{
char data;
struct node *nextPtr;
}*LinkList, Lnode;
static void CreateList(LinkList *headPtr, LinkList *tailPtr, char ch);
static void Decompose(LinkList *headPtrA, LinkList *headPtrB, LinkList *tailPtrB);
static void VisitList(LinkList headPtr);
static void DestroyList(LinkList *headPtr, LinkList *tailPtr);
int main(void)
{
LinkList headPtrA = NULL, tailPtrA = NULL, headPtrB = NULL, tailPtrB = NULL;
char ch;
while (1)
{
printf("Enter ch('@'-quit): ");
scanf(" %c", &ch);
if (ch == '@')
{
break;
}
else
{
CreateList(&headPtrA, &tailPtrA, ch);
}
}
VisitList(headPtrA); /* \u6253\u5370\u5206\u89e3\u524d\u7684\u94fe\u8868 */
if (headPtrA != NULL) /* \u94fe\u8868\u4e0d\u7a7a\u7684\u60c5\u51b5\u5bf9\u5176\u8fdb\u884c\u5206\u89e3 */
{
Decompose(&headPtrA, &headPtrB, &tailPtrB); /* \u5bf9\u94fe\u8868\u8fdb\u884c\u5206\u89e3*/
}
else
{
printf("headPtrA is empty.\n");
}
VisitList(headPtrA); /* \u6253\u5370\u5206\u89e3\u540e\u7684\u94fe\u8868 */
VisitList(headPtrB);
DestroyList(&headPtrA, &tailPtrA); /* \u9500\u6bc1\u94fe\u8868 */
DestroyList(&headPtrB, &tailPtrB);
return 0;
}
static void CreateList(LinkList *headPtr, LinkList *tailPtr, char ch)
{
LinkList newPtr;
if ((newPtr = (LinkList)malloc(sizeof(Lnode))) == NULL)
{
exit(1);
}
newPtr -> data = ch;
newPtr -> nextPtr = NULL;
if (*headPtr == NULL)
{
newPtr -> nextPtr = *headPtr;
*headPtr = newPtr;
}
else
{
(*tailPtr) -> nextPtr = newPtr;
}
*tailPtr = newPtr;
}
static void Decompose(LinkList *headPtrA, LinkList *headPtrB, LinkList *tailPtrB)
{
int count = 0;
LinkList cA, pA;
char ch;
cA = NULL;
for (pA = *headPtrA; pA != NULL; cA = pA,pA = pA -> nextPtr)
{
ch = pA -> data;
count++;
if (count % 2 == 0)
{
CreateList(headPtrB, tailPtrB, ch);
cA -> nextPtr = pA -> nextPtr;
}
}
}
static void VisitList(LinkList headPtr)
{
while (headPtr != NULL)
{
printf("%c -> ", headPtr -> data);
headPtr = headPtr -> nextPtr;
}
printf("NULL\n");
}
static void DestroyList(LinkList *headPtr, LinkList *tailPtr)
{
LinkList tempPtr;
while (*headPtr != NULL)
{
tempPtr = *headPtr;
*headPtr = (*headPtr) -> nextPtr;
free(tempPtr);
}
*headPtr = NULL;
*tailPtr = NULL;
}
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1986\u5e7410\u6708\uff0c\u7f8e\u56fd\u56fd\u5bb6\u6807\u51c6\u534f\u4f1a\u5bf9SQL\u8fdb\u884c\u89c4\u8303\u540e\uff0c\u4ee5\u6b64\u4f5c\u4e3a\u5173\u7cfb\u5f0f\u6570\u636e\u5e93\u7ba1\u7406\u7cfb\u7edf\u7684\u6807\u51c6\u8bed\u8a00\uff08ANSI X3. 135-1986\uff09\uff0c1987\u5e74\u5f97\u5230\u56fd\u9645\u6807\u51c6\u7ec4\u7ec7\u7684\u652f\u6301\u4e0b\u6210\u4e3a\u56fd\u9645\u6807\u51c6\u3002\u4e0d\u8fc7\u5404\u79cd\u901a\u884c\u7684\u6570\u636e\u5e93\u7cfb\u7edf\u5728\u5176\u5b9e\u8df5\u8fc7\u7a0b\u4e2d\u90fd\u5bf9SQL\u89c4\u8303\u4f5c\u4e86\u67d0\u4e9b\u7f16\u6539\u548c\u6269\u5145\u3002\u6240\u4ee5\uff0c\u5b9e\u9645\u4e0a\u4e0d\u540c\u6570\u636e\u5e93\u7cfb\u7edf\u4e4b\u95f4\u7684SQL\u4e0d\u80fd\u5b8c\u5168\u76f8\u4e92\u901a\u7528\u3002
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楼上的说了原理，我还是给一个运行通过的程序你吧。遍历La，大于等于0放在Lb，其他放在Lc。
#include "stdafx.h"
#include <stdlib.h>
#include <stdio.h>
typedef struct NODE
{
struct NODE *next;
int data;
} T_NODE;
// 输入链表存储
void InputData(struct NODE * pHead)
{
int count = 10;
T_NODE *p = pHead;
while(count-- > 0)
{
p->next = (struct NODE *)malloc(sizeof(struct NODE));
scanf("%d", &(p->next->data));
p->next->next = NULL;
p = p->next;
}
p = pHead;
while(p->next)
{
printf("%d ", p->next->data);
p = p->next;
}
printf("\n");
}
int _tmain(int argc, _TCHAR* argv[])
{
// 生成单链表头结点
struct NODE * pHead_La = (struct NODE *)malloc(sizeof(struct NODE));
pHead_La->data = 0;
pHead_La->next = NULL;
// 输入正负数
InputData(pHead_La);
struct NODE * pHead_Lb = (struct NODE *)malloc(sizeof(struct NODE));
pHead_Lb->data = 0;
pHead_Lb->next = NULL;
struct NODE * pHead_Lc = (struct NODE *)malloc(sizeof(struct NODE));
pHead_Lc->data = 0;
pHead_Lc->next = NULL;
T_NODE *p = pHead_La;
T_NODE *pPostive = pHead_Lb;
T_NODE *pNegative = pHead_Lc;
while(p->next)
{
// 正数和0
if (p->next->data >= 0)
{
pPostive->next = (struct NODE *)malloc(sizeof(struct NODE));
pPostive->next->data = p->next->data;
pPostive->next->next = NULL;
pPostive = pPostive->next;
}
else
{
pNegative->next = (struct NODE *)malloc(sizeof(struct NODE));
pNegative->next->data = p->next->data;
pNegative->next->next = NULL;
pNegative = pNegative->next;
}
p = p->next;
}
pPostive = pHead_Lb;
while(pPostive->next)
{
printf("%d ", pPostive->next->data);
pPostive = pPostive->next;
}
printf("\n");
pNegative = pHead_Lc;
while(pNegative->next)
{
printf("%d ", pNegative->next->data);
pNegative = pNegative->next;
}
printf("\n");
return 0;
}
遍历La，如果是负整数就接在Lb后边，要不然就接在Lc后边呗
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绛旓細include <stdio.h> void delRe(int a[],int &length) //length涓鸿繑鍥炴暟缁勯暱搴 {int i,j,t;t=length;i=0; //i鎸囧悜鏃犻噸澶嶅尯鍩熺殑鏈熬 j=1; //j涓鸿妫鏌ョ殑鍏冪礌 while(j<t)if(a[i]==a[j]){j++;length--;} else {i++;a[i]=a[j];j++;} } main(){int i,n,ar...
