[编程题]合并两个排序的链表

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。

递归方法：

/*
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;}
        if(pHead1->val<=pHead2->val)
        {
            pHead1->next=Merge(pHead1->next, pHead2);
            return pHead1;
        }
        else
        {
            pHead2->next=Merge(pHead1,pHead2->next);
            return pHead2;
        }
    }
};

非递归算法

/*
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* list=NULL;
        ListNode* current=NULL;
        while(pHead1!=NULL&&pHead2!=NULL)
        {
            if(pHead1->val<=pHead2->val)
        {
           if(list==NULL)
           {
               list=current=pHead1;
           } 
           else 
           {
               current->next=pHead1;
               current=current->next;
           }
           pHead1=pHead1->next; 
        }
        else
        {
            if(current==NULL)
           {
               list=current=pHead2;
           } 
           else 
           {
               current->next=pHead2;
               current=current->next;
           }
           pHead2=pHead2->next;             
        }
       }
        if(pHead1==NULL){current->next=pHead2;}
        else if(pHead2==NULL){current->next=pHead1;}
        return list;
    }
};