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。

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
//方法一：非递归
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        //创建一个结点为0的链表
        ListNode* current = new ListNode(0);
        ListNode* result= current;
        //开始遍历两个链表，将较小当前较小值的结点添加进current链表中
        while (pHead1 != NULL && pHead2 != NULL)
        {
            if (pHead1->val <= pHead2->val) {
                current->next = pHead1;
                current = current->next;
                pHead1 = pHead1->next;
                
            }
            else
            {
                current->next= pHead2;
                current = current->next;
                pHead2 = pHead2->next;
                
            }
        }
        //那个链表没遍历完，直接添加未遍历完的链表
        if (pHead1 != NULL) {
            current->next = pHead1;
        }
        if (pHead2 != NULL) {
            current->next = pHead2;
        }
        //返回合并后的链表
        return result->next;
    }
};

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
//方法二：递归
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        if (pHead1==NULL) {
            return pHead2;
        }
        if (pHead2 == NULL) {
            return pHead1;
        }
        ListNode * result = NULL;
        if (pHead1->val <= pHead2->val) {
            result = pHead1;
            result->next = Merge(pHead1->next,pHead2);
        }
        else
        {
            result = pHead2;
            result->next = Merge(pHead1, pHead2->next);
        }
        return result;
    }
};