leetcode 题解Merge Two Sorted Lists（有序链表归并）

题目：

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

说明：有序链表归并（从小到大）

       1）此链表无头节点

实现：

       方法一：非递归    

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode *la=l1,*lb=l2;
        ListNode *q=NULL,*p=NULL;
        if(la==NULL) return l2;
        if(lb==NULL) return l1;
        //此单链表无头结点，若有头结点，可直接p=head节点即可，无需下面的if...else...
        if((la->val) < (lb->val))
            {
                p=la;
                la=la->next;
            }
        else
            {
                p=lb;
                lb=lb->next;
            }
        q=p;
        while(la&&lb)
        {
            if((la->val) < (lb->val))
            {
                p->next=la;
                p=la;
                la=la->next;
            }
            else
            {
                p->next=lb;
                p=lb;
                lb=lb->next;
            }
        }
        p->next=la?la:lb;
        return q;
    }
};

       方法二：递归（紫红的泪大牛的代码，博客链接：http://www.cnblogs.com/codingmylife/archive/2012/09/27/2705286.html）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if (l1 == NULL) return l2;
    if (l2 == NULL) return l1;
    
    ListNode *ret = NULL;
    
    if (l1->val < l2->val)
    {
        ret = l1;
        ret->next = mergeTwoLists(l1->next, l2);
    }
    else
    {
        ret = l2;
        ret->next = mergeTwoLists(l1, l2->next);
    }
    
    return ret;
    }
};