标签：sorted list 情况 else font 指针 sort new while
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.
题解：分情况考虑：当两个链表有一个为空的时候，返回另一个链表即可；
当两个链表都不为空时，依次进行比较l1和l2的值，直至某一链表指针为空，之后直接将另一链表的剩余部分插入链表尾部即可。
/** * 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* head=new ListNode(0); ListNode* ret=head; if(l1==NULL&&l2!=NULL) return l2; if(l1!=NULL&&l2==NULL) return l1; while(l1!=NULL&&l2!=NULL) { if(l1->val<=l2->val) { head->next=l1; l1=l1->next; } else { head->next=l2; l2=l2->next; } head=head->next; } while(l1) { head->next=l1; l1=l1->next; head=head->next; } while(l2) { head->next=l2; l2=l2->next; head=head->next; } return ret->next; } };
Leetcode-21 Merge Two Sorted Lists
标签：sorted list 情况 else font 指针 sort new while
原文地址：http://www.cnblogs.com/fengxw/p/6082852.html
