Merge Two Sorted Lists——解题报告

标签：leetcode   链表   合并   递归   指针   

    【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;
        
    }
};

Merge Two Sorted Lists——解题报告

标签：leetcode   链表   合并   递归   指针   

原文地址：http://blog.csdn.net/puqutogether/article/details/45578391