Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
C++代码如下:
#include<iostream> #include<new> using namespace std; //Definition for singly-linked list. struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { ListNode *p=NULL; //q始终记录l2中的元素,qq是取出来要插入到l1中的元素 ListNode *q=NULL; ListNode *qq=NULL; //pre是p的前驱,插入比p小的元素时需要 ListNode *pre=NULL; if(l1&&l2) { pre=p=l1; qq=q=l2; while(p&&q) { if(p->val<=q->val) { pre=p; p=p->next; } else { qq=q; q=q->next; if(l1==p) { qq->next=l1; l1=qq; pre=p=l1; continue; } qq->next=p; pre->next=qq; pre=qq; } } while(q) { pre->next=q; pre=q; q=q->next; } } if(l1==NULL) l1=l2; return l1; } void createList(ListNode *&head,int *arr) { ListNode *p=NULL; int i=0; for(i=0; i<5; i++) { if(head==NULL) { head=new ListNode(arr[i]); if(head==NULL) return; } else { p=new ListNode(arr[i]); p->next=head; head=p; } } } }; int main() { Solution s; ListNode *L1=NULL; ListNode *L2=NULL; ListNode *L=NULL; int arr1[10]= {11,9,7,5,3}; int arr2[10]= {10,8,6,4,2}; s.createList(L1,arr1); s.createList(L2,arr2); L=s.mergeTwoLists(L1,L2); while(L) { cout<<L->val<<" "; L=L->next; } }
运行结果:
第二遍:
class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { if(l1==NULL) return l2; if(l2==NULL) return l1; ListNode *head=NULL; ListNode *r=NULL; ListNode *p=l1; ListNode *q=l2; if(p->val<q->val) { head=p; r=head; p=p->next; } else { head=q; r=head; q=q->next; } r->next=NULL; while(p&&q) { if(p->val<q->val) { r->next=p; r=p; p=p->next; } else { r->next=q; r=q; q=q->next; } r->next=NULL; } if(q) p=q; r->next=p; return head; } };
时间: 2024-09-26 14:29:30