【数据结构·C语言】请高手帮忙检查一个关于【链表归并】的算法是否正确 【数据结构·C语言】请高手帮忙检查一个关于【实现链表的就地逆...
\u3010\u6570\u636e\u7ed3\u6784\u00b7C\u8bed\u8a00\u3011\u8bf7\u9ad8\u624b\u5e2e\u5fd9\u68c0\u67e5\u4e00\u4e2a\u5173\u4e8e\u3010\u94fe\u8868\u7684\u5f52\u5e76\u3011\u7b97\u6cd5\u662f\u5426\u6b63\u786efor(p=&(A.elem[A.length-1]);p>=&(A.elem[i]);p--) *(p+1)=p; \u6700\u540e\u7684p\u524d\u9762\u5c11\u4e00\u4e2a\u661f\u53f7\uff0c\u5e94\u8be5\u6539\u4e3afor(p=&(A.elem[A.length-1]);p>=&(A.elem[i]);p--) *(p+1)=*p;
A.elem[i]=B.elem[k];\u4e4b\u540e\uff0c\u7531\u4e8eB\u7684\u5143\u7d20\u63d2\u5728\u4e86A\u7684i\u4f4d\u7f6e\uff0c\u6240\u4ee5\uff0c\u8fd9\u4e4b\u540ei\u5e94\u8be5\u81ea\u52a01
for (p=&(A.elem[i+1]);p<=&(A.elem[A.length-1]);p++) *(p-1)=*p;\u8fd9\u91cc\u662f\u8981\u8986\u76d6i+1\u4f4d\u7f6e\u7684\u5143\u7d20\uff0c\u800c\u6839\u636e\u4f60\u81ea\u5df1\u5199\u7684\u521d\u59cb\u6761\u4ef6\uff0cp\u4e00\u5f00\u59cb\u5c31\u5df2\u7ecf\u6307\u5411i+1\u4f4d\u7f6e\uff0c\u56e0\u6b64\u9700\u8981\u4fee\u6539\uff0c\u540c\u65f6\u4f34\u968f\u4fee\u6539\u7ed3\u675f\u6761\u4ef6\uff0c\u6700\u7ec8\u4fee\u6539\u4e3afor (p=&(A.elem[i+1]);p<&(A.elem[A.length-1]);p++) *p=*(p+1);
\u5176\u4ed6\u7684\u90fd\u8fd8\u597d\u3002
PS\uff0c\u5982\u679c\u60f3\u7a0b\u5e8f\u8fd0\u884c\u7b80\u5355\u4e00\u70b9\uff0c\u5efa\u8bae\u4f60\u6bcf\u5f52\u5e76\u4e00\u4e2aB\u7684\u5143\u7d20\u5230A\u5c31\u67e5\u627e\u4e00\u904dA\u4e2d\u7684\u5143\u7d20\uff0c\u4f46\u662f\u7a0b\u5e8f\u5c31\u4f1a\u9ebb\u70e6\u4e00\u70b9\uff0c\u8fd8\u6709\uff0c\u5bf9A\u7684\u91cd\u590d\u5143\u7d20\u7684\u81ea\u68c0\uff0c\u53ef\u4ee5\u8003\u8651\u628aif\u5199\u4e3awhile\uff0c\u56e0\u4e3a\u8003\u8651\u5230\u53c8\u53ef\u4ee5\u80fd\u8fde\u7740\u4e09\u4e2a\u4ee5\u4e0a\u7684\u5143\u7d20\u76f8\u7b49\u3002
\u5e0c\u671b\u91c7\u7eb3\uff0c\u671f\u5f85\u5bf9\u4f60\u6709\u5e2e\u52a9\uff0c\u6b22\u8fce\u8ffd\u95ee^_^
\u6211\u5df2\u7ecf\u628a\u4f60\u7ed9\u7684\u4ee3\u7801\u5728\u4e0b\u5217\u4ee3\u7801\u4e2d\u505a\u4e86\u5177\u4f53\u5b9e\u73b0\uff0c\u5177\u4f53\u95ee\u9898\u5df2\u5728\u4e0b\u5217\u4ee3\u7801\u4e2d\u8bf4\u660e
#include
using namespace std;
#define OK true
#define FALSE false
typedef bool status;
typedef int INT32;
typedef struct tagLinkNode
{
INT32 key;
struct tagLinkNode *next;
tagLinkNode()//construct
{
key = -1;
next = NULL;
}
}LinkNode, *LinkList;
status createLinkList( LinkList &L)
{
INT32 nAccount = 0 ;//number of node in LinkList L
cout<<"input the num of node,please"<<endl;
cin>>nAccount;
//
//the first node is empty
//
L = ( LinkList ) malloc( sizeof( LinkNode) );
L->next = NULL;
LinkList pNow = L;
//create list
cout<<"import data:"<<endl;
for( INT32 i = 0; i<nAccount; i++ )
{
LinkList pTemp = ( LinkList ) malloc( nAccount * sizeof( LinkNode ) );
if( pTemp == NULL ) return FALSE;
pTemp->next = NULL;
INT32 nKey;
cin>>nKey;
pTemp->key = nKey;
pNow->next = pTemp;
pNow = pTemp;
}
return OK;
}
void clearLinkList( LinkList &L )
{
LinkList pNow = L;
while( pNow != NULL )
{
LinkList pTemp = pNow->next;
free( pNow );
pNow = pTemp;
}
L = NULL;
return ;
}
status inverseLinkList( LinkList &L )//\u8fd9\u91cc\u7684status\u8bf4\u660e\u4f60\u7684\u4ee3\u7801\u5e94\u8be5\u662f\u4eff\u7167\u4e66\u6765\u5199\u7684\uff0c\u4e66\u4e0a\u7684\u4ee3\u7801\u4e00\u822c\u662f\u4f2a\u7801
{
LinkList pNow = L,pTemp = L;// \u4f60\u6ca1\u6709\u8d4b\u503c
while( pNow != NULL )//traverse in first loop
{
//
//\u4e0b\u9762\u8fd9\u53e5\u662f\u4e0d\u5b89\u5168\u7684\uff0c\u4f60\u6ca1\u6709\u5224\u65adpTemp->next\u7684\u6709\u6548\u6027
//
/*for( ; pTemp->next->next != NULL; pTemp = pTemp->next );*/
if( pTemp->next == NULL ) break;
if( pTemp->next->next == NULL )// \u907f\u514d\u7b2c\u4e00\u4e2a\u8282\u70b9\u4e22\u5931
{
break;
}
for( ; pTemp->next->next != NULL; pTemp = pTemp->next );//traverse in second loop \u5f88\u663e\u7136\u7b97\u6cd5\u7684\u65f6\u95f4\u590d\u6742\u5ea6\u4e3a0(n^2)
//
//\u4e0b\u9762\u4e09\u884c\u4ee3\u7801\u6709\u95ee\u9898\u7684\uff0c\u4f1a\u4e22\u5931\u539f\u94fe\u8868\u4e2d\u7684\u7b2c\u4e00\u4e2a\u8282\u70b9
//
pTemp->next->next = pNow->next;
pNow->next = pTemp->next;
pTemp->next = NULL;
pNow = pNow->next;
pTemp = pNow;
}
return OK;
}
void displayLinkList( LinkList L )
{
LinkList pNow = L->next;
cout<<"print LinkList:"<<endl;
while( pNow != NULL )
{
coutkey <<" ";
pNow = pNow->next;
}
cout<<endl;
return ;
}
int main()
{
LinkList L = NULL;
createLinkList( L );// create LinkList
cout<<"before the operation of inversing"<<endl;
displayLinkList( L );
inverseLinkList( L );
cout<<"after the operation of inversing"<<endl;
displayLinkList( L );
clearLinkList( L );
system("pause");
return 0;
}
2. 您的算法因为AB是递增有序要改为递减有序，您就每次将指针移动到序列的最末端来进行比较和插入，由于是单向链表，这样你的算法会非常低效。
3. 您仅仅需要将两个链表的结点按照递减顺序插入到新的链表中即可。
4. 例如您可以先将A中的结点逆序，然后将B中的结点按照递减顺序一个一个插入到A中的合适位置，最终获得的A链表即为需要的C链表了。
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