剑指offer 16：合并两个排序的链表

十六题目描述
输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。
思路：设置两个指针分别指向这两个递增链表。存储对应小的数，并对应的指针自加。当其中一个链表为空时，则把新链表指向另一个链表。

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        ListNode *p1=pHead1, *p2=pHead2, *head=new ListNode(0);
        head->next=NULL;
        ListNode* p0=head;
        while(p1 !=NULL && p2 !=NULL)
        {
            if(p1->val > p2->val)
            {
                p0->next=p2;
                p2=p2->next;
            }
            else
            {
                p0->next=p1;
                p1=p1->next;
            }
            p0=p0->next;
        }
        if(p1==NULL)
            p0->next=p2;
        else
            p0->next=p1;
        return head->next;
    }
};
/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        if(pHead1==NULL)
            return pHead2;
        if(pHead2==NULL)
            return pHead1;
        ListNode* head=NULL;
        if(pHead1->val <pHead2->val)
        {
            head=pHead1;
            head->next=Merge(pHead1->next,pHead2);
        }
        if(pHead1->val >pHead2->val)
        {
            head=pHead2;
            head->next=Merge(pHead1,pHead2->next);
        }
        return head;

    }
};
# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        p1=pHead1
        p2=pHead2
        head=ListNode(0)
        head.next=None
        p0=head
        while p1!=None and p2!=None:
            if p1.val>p2.val:
                p0.next=p2
                p2=p2.next
            else:
                p0.next=p1
                p1=p1.next
            p0=p0.next
        if p1==None:
            p0.next=p2
        else:
            p0.next=p1
        return head.next
        # write code here