Merge Two Sorted Lists
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.
思路:1. 一般做法。用两个指针分别指两条链表,一个一个比较。
2.用一个指针,将整个链表按大小串起来。
注意“等号”用在何时!!!容易错.
/** * 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) { if(l1==NULL) return l2; else if(l2==NULL) return l1; ListNode *p1=l1; ListNode *p2=l2; ListNode *head,*t; if(l1->val<=l2->val) head=l1; else head=l2; while(p1 && p2 ){ if(p1->val<=p2->val){ if(p1->next && p1->next->val<=p2->val) p1=p1->next; else { t=p1; p1=p1->next; t->next=p2; } }else{ if(p2->next && p2->next->val<p1->val) p2=p2->next; else{ t=p2; p2=p2->next; t->next=p1; } } } return head; } };
2.代码更简洁!!!
/** * 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) { if(l1==NULL) return l2; if(l2==NULL) return l1; ListNode *head,*t; if(l1->val<=l2->val) {head=l1;l1=l1->next;} else {head=l2;l2=l2->next;} t=head; while(l1 && l2){ if(l1->val<=l2->val) {t->next=l1;l1=l1->next;} else {t->next=l2;l2=l2->next;} t=t->next; } if(l1) t->next=l1; else t->next=l2; return head; } };
turnsoul