剑指offer--合并两个排序的链表

public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
public class Solution {

    public ListNode Merge(ListNode list1,ListNode list2) {
        ListNode list = new ListNode(0);
        ListNode cur = list;
        if ( list1 == null){
            return list2;
        }

        if (list2 == null){
            return list1;
        }

        while(list1 != null & list2 != null){
            if ( list1.val <= list2.val){
                list.next = list1;
                list = list.next;
                list1 = list1.next;
            }else{
                list.next = list2;
                list = list.next;
                list2 = list2.next;
            }
        }

        while(list1 != null){
            list.next = list1;
            list = list.next;
            list1 = list1.next;
        }

        while(list2 != null){
            list.next = list2;
            list = list.next;
            list2 = list2.next;
        }

        list.next = null;
        cur = cur.next;

        return cur;
    }
}

struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};
class Solution {
    public:
        ListNode* Merge(ListNode* pHead1, ListNode* pHead2){
            ListNode* head = new ListNode(0);
            ListNode* tmp = head;
            ListNode* p1 = pHead1, *p2 = pHead2;
            while(p1 && p2){
                ListNode* p;
                if(p1->val <= p2->val){
                    p = new ListNode(p1->val);
                    p1 = p1->next;
                }else{
                    p = new ListNode(p2->val);
                    p2 = p2->next;
                }
                tmp->next = p;
                tmp = tmp->next;
            }
            if(p1){
                tmp->next = p1;
            }
            if(p2){
                tmp->next = p2;
            }
            head = head->next;
            return head;
        }
};