LeetCode—Merge Two Sorted Lists融合两个有序单链表

首先两个单链表是有序的

在融合两个单链表的时候，如果想到的是在一个序列上进行增减，那么会非常麻烦

这里一定要单独开一个序列头进行存储，不一定需要开辟内存，主要是一个概念

其实方法感觉和归并算法的merge都是一个概念

/**
 * 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;
        }
        else if(l2 == NULL)
        {
            return l1;
        }
        ListNode *head = NULL;
        if(l1->val < l2->val)
        {
            head = l1;
            l1 = l1->next;
        }
        else
        {
            head = l2;
            l2 = l2->next;
        }
        ListNode *phead = head;
        while(l1 && l2)
        {
            if(l1->val < l2->val)
            {
                head->next = l1;
                head = head->next;
                l1 = l1->next;
            }
            else
            {
                head->next = l2;
                head = head->next;
                l2 = l2->next;
            }
        }
        if(l1)
        {
            head->next = l1;
        }
        if(l2)
        {
            head->next = l2;
        }
        return phead;
        
    }
};

提供剑指offer上的递归的解决办法

/**
 * 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;
        }
        else if(l2 == NULL)
        {
            return l1;
        }
        ListNode *head = NULL;
        if(l1->val < l2->val)
        {
            head = l1;
            head->next = mergeTwoLists(l1->next,l2);
        }
        else
        {
            head = l2;
            head->next = mergeTwoLists(l1,l2->next);
        }
        return head;
    }
};