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.
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *mergeTwoLists(ListNode *list1, ListNode *list2) { // Start typing your C/C++ solution below // DO NOT write int main() function if(list1 == NULL && list2 == NULL) return NULL; ListNode *head = NULL, *cur = NULL; while(list1 && list2){ ListNode *tp; if(list1->val >list2->val){ tp = list2; list2 = list2->next; }else{ tp = list1; list1 = list1->next; } if(cur == NULL){ head = cur = tp; }else{ cur->next = tp; cur = tp; } } list1 = list1 == NULL ? list2 : list1; if(cur == NULL) return list1; else cur->next = list1; return head; } };
重写后: 定义一个伪头,省去所有的边界问题。
/** * 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) { // Start typing your C/C++ solution below // DO NOT write int main() function if(l1 == NULL) return l2; if(l2 == NULL) return l1; ListNode * head = new ListNode(1); ListNode * p = head; while(l1 != NULL && l2 != NULL){ if(l1->val < l2->val){ p->next = l1; p = l1; l1 = l1->next; }else{ p->next = l2; p = l2; l2 = l2->next; } } p->next = l1 == NULL ? l2 : l1; p = head->next; delete head; return p; } };