LeetCode_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.

　　

/**

 * 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;    

    }

};