正文
LeetCode OJ--Merge Two Sorted Lists
提示：扫一扫查出行【扫一扫了解最新限行尾号】
复制提示
http://oj.leetcode.com/problems/merge-two-sorted-lists/
有序链表的归并排序
#include <iostream>
using namespace std; struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
}; class Solution {
public:
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
// default as asend sorted,merge sort
ListNode *ans = NULL,*ansEnd = NULL,*n1 = l1,*n2 = l2;
bool firstone = ;
while(n1&&n2)
{
while(n1->val <= n2 ->val)
{
if(firstone == )
{
ans = ansEnd = n1;
//ansEnd = ans->next;
firstone = ;
n1 = n1->next;
}
else
{
ansEnd->next = n1;
ansEnd = n1;
n1 = n1->next;
}
if(n1 == NULL)
break;
}
if(n1 == NULL)
break;
while(n1->val > n2->val)
{
if(firstone == )
{
ans = ansEnd = n2;
firstone = ;
n2 = n2->next;
}
else
{
ansEnd->next = n2;
ansEnd = n2;
n2 = n2->next;
}
if(n2 == NULL)
break;
}
}
if(n1==NULL && n2!= NULL)
{
if(firstone ==)
{
ans = n2;
return ans;
}
ansEnd->next = n2;
}
else if(n2 == NULL && n1 != NULL)
{
if(firstone == )
{
ans = n1;
return ans;
}
ansEnd->next = n1;
}
return ans;
}
}; int main()
{
ListNode *n1 = new ListNode();
ListNode *n2 = new ListNode();
ListNode *n3 = new ListNode(); n1->next = n2; Solution myS;
ListNode *ans = myS.mergeTwoLists(NULL,NULL);
return ;
}
