17 合并两个排序的链表

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。

C++：迭代

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

C++：递归

/*
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 ;
        }
    }
};