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description:
given two sorted singly list, merge them into one using constant additional space
algorithm:
we will reference the two linked list as list1 and list2 for convenience,
since list1 is sorted,just find the right position for each element in list2,
detailed comments are added to the following code
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#include<iostream>#include<cstdio>#include<string.h>#include<string>#include<cstring>#include<algorithm>using
namespace std;class
Node{public: int
value; Node* Next; Node(int
v_ = 0, Node* Next_ = NULL):value(v_),Next(Next_){}};/*Node* Merge2(Node* head1, Node* head2){ Node* res,*ret; if(head1 == NULL) return head2; if(head2 == NULL) return head1; Node* p = head1; Node* q = head2; if(p->value < q->value) { res = p; p = p->Next; } else { res = q; q = q->Next; } ret = res; while(p && q) { if(p->value < q->value) { res->Next = p; res = p; p = p->Next; } else { res->Next = q; res = q; q = q->Next; } } while(p) { res->Next = p; res = p; p = p->Next; } while(q) { res->Next = q; res = q; q = q->Next; } return ret;}*/Node* Merge2(Node*p1,Node*p2){ if(p1 == NULL) return
p2; if(p2 == NULL) return
p1; Node *tmpp1 = p1,*tmpp2 = p2; Node*head; bool
first = true; /*core optimization: if tmpp2 find its place in list1, then tmpp2->next must be after its position,so both list1 and list2 are enumerated only once, which implements linear algorithm*/ while(tmpp1 != NULL && tmpp2 != NULL){ Node* tmpformer = NULL; while(tmpp1 != NULL && tmpp2->value > tmpp1->value){ tmpformer = tmpp1; tmpp1 = tmpp1->Next; } if(tmpp1 == NULL){ tmpformer->Next = tmpp2; if(first){ first = false; head = p1; } break; } //tmpp2的value值比tmpp1的value值小但是比tmpformer的value值大 Node* tmprecordp2 = tmpp2->Next; tmpp2->Next = tmpp1; if(tmpformer != NULL){ tmpformer->Next = tmpp2; if(first){ first = false; head = p1; } } else{ if(first){ first = false; head = p2; } } tmpp2 = tmprecordp2; } return
head;}int
main(){ Node* p1[10], *p2[10]; memset(p1,0,sizeof(p1)); memset(p2,0,sizeof(p2)); int
a[25] = {1,3,6,7,8,9,56,67,211,763,2,4,5,10,11,12,13,300,500,800}; /*initalization*/ for(int
i = 0; i < 10; ++i){ p1[i] = new
Node(a[i]); } for(int
i = 0; i < 10; ++i){ p2[i] = new
Node(a[10+i]); } Node* pp1,*pp2; for(int
i = 0; i < 9; ++i){ // cout << &(p1[i]->Next) << endl; p1[i]->Next = p1[i+1]; } for(int
i = 0; i < 9; ++i){ p2[i]->Next = p2[i+1]; } pp1 = p1[0]; while(pp1 != NULL){ cout << pp1->value << ‘\t‘; pp1 = pp1->Next; } cout << endl; pp2 = p2[0]; while(pp2 != NULL){ cout << pp2->value << ‘\t‘; pp2 = pp2->Next; } cout << endl; /*initialization end*/ Node* res = Merge2(p1[0],p2[0]); while(res!=NULL){ cout << res->value << endl; res = res->Next; } for(int
i = 0; i < 10; ++i){ delete
p1[i]; } for(int
i = 0; i < 10; ++i){ delete
p2[i]; } return
0;} |
key points:
1.pay special attention to keep head pointer, which is easily lost,
my solution is to set a flag bool variance, which is linear complexity but somewhat inefficient
perhaps it has potential to improve
2.avoid misorder of pointer
3.special condition: if (p1 == NULL) return p2;
if(p2 == NULL ) return p1;
4.when search for the postion for a certain element in list1, note that it might be above any element in list1 so tmpp1 might reach its end(value equals NULL) during search
线性时间将两个有序链表合成一个有序链表(constant additional space),码迷,mamicode.com
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