Merge Two Sorted Lists——解题报告

【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.

【分析】

不要忘记先判断两个链表是否有空链表。其余的使用递归和非递归方式，都可以实现。

【代码】

递归方式：

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* mergedList = NULL; 

        if(l1 == NULL)
            return l2;
        if(l2 == NULL)
            return l1;

        if(l1->val < l2->val)
        {
            mergedList = l1;
            mergedList->next = mergeTwoLists(l1->next, l2);
        }
        else
        {
            mergedList = l2;
            mergedList->next = mergeTwoLists(l1, l2->next);
        }

        return mergedList;

    }
};

非递归方式：

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* mergedList = NULL; 

        if(l1 == NULL)
            return l2;
        if(l2 == NULL)
            return l1;

        if(l1->val < l2->val)
        {
            mergedList = l1;
            mergedList->next = NULL;
            l1 = l1->next;
        }
        else
        {
            mergedList = l2;
            mergedList->next = NULL;
            l2 = l2->next;
        }

        ListNode* p = mergedList;
        while(l1 != NULL && l2 != NULL)
        {
            if(l1->val < l2->val)
            {
                p->next = l1;
                l1 = l1->next;
                p = p->next;
                p->next = NULL;
            }
            else
            {
                p->next = l2;
                l2 = l2->next;
                p = p->next;
                p->next = NULL;
            }
        }

        if(l1 != NULL)
            p->next = l1;
        else if(l2 != NULL)
            p->next = l2;

        return mergedList;

    }
};