最后合并的C也要有序，并且不能破坏A,B ```cpp #include #include using namespace std; typedef struct LNode { int data; struct LNode* next; }LinkNode; void Create(LinkNode*& L, int a[], int l) { L = (LinkNode*)malloc(sizeof(LinkNode)); LinkNode* q = L; for (int i = 0; i < l; i++) { LinkNode* p = (LinkNode*)malloc(sizeof(LinkNode)); p->data = a[i]; q->next = p; q = p; } q->next = NULL; } void Union2(LinkNode* A, LinkNode* B, LinkNode*& C) { LinkNode* pa, * pb, * pc,*p; C= (LinkNode*)malloc(sizeof(LinkNode)); pc = C; pa = A->next; pb = B->next;//因为建表都是有建头结点的 while (pa != NULL && pb != NULL) { p = (LinkNode*)malloc(sizeof(LinkNode)); if (pa->data > pb->data) { p->data = pb->data; pb = pb->next; } else if(pa->data< pb->data){ p->data = pa->data; pa = pa->next; } else { p->data = pa->data; pa = pa->next; pb = pb->next; } pc->next = p; pc = p; } while (pa != NULL) { p= (LinkNode*)malloc(sizeof(LinkNode)); p->data = pa->data; pc->next = p; pc = p; pa = pa->next; } while (pb != NULL) { p = (LinkNode*)malloc(sizeof(LinkNode)); p->data = pb->data; pc->next = p; pc = p; pb = pb->next; } pc -> next = NULL; } void display(LinkNode* L) { LinkNode* p = L->next; while (p != NULL) { cout << p->data << " "; p = p->next; } } int main() { int a1[] = { 1,2,3,4,5 }, b1[] = { 4,5,6 }; LinkNode* A, * B, * C; Create(A, a1, 4); display(A); cout << endl; Create(B, b1, 3); display(B); cout << endl; Union2(A, B, C); display(C); return 0; } ``` 细节在上一篇博客里；有序合并的思路就就是，同时循环两个表比较，最后在判断哪个表没空就把剩余的加进去