Merge Two Sorted Lists

最新推荐文章于 2022-05-22 13:07:29 发布

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最新推荐文章于 2022-05-22 13:07:29 发布

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原文链接：http://www.cnblogs.com/pengyu2003/p/3572425.html

本文介绍了一种合并两个已排序的链表的方法，并提供了一个C++实现示例。该算法通过比较两个链表节点的值来决定合并顺序，最终返回一个新的合并后的有序链表。

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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.

if... else if不要偷懒直接写 if...if...

/**
 * 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 *head = NULL,*temp = head;
        
        if(l1 == NULL && l2 == NULL)return NULL;
        if(l1 == NULL && l2 != NULL)return l2;
        if(l1 != NULL && l2 == NULL)return l1;
        
        while(l1 != NULL && l2 != NULL)
        {
            if(l1->val <= l2->val)
            {
                if(head == NULL)
                {
                    head = l1;
                    temp = head;
                    l1 = l1->next;
                    continue;
                }
                
                temp->next = l1;
                l1 = l1->next;
                temp = temp->next;
                
                
            }
            else if(l1->val > l2->val)
            {
                if(head == NULL)
                {
                    head = l2;
                    temp = head;
                    l2 = l2->next;
                    continue;
                }
                
                temp->next = l2;
                l2 = l2->next;
                temp = temp->next;
                
                
            }
        }
        if(l1 == NULL)temp ->next = l2;
        else if(l2 == NULL)
        temp ->next =l1;
        
        return head;
    }
};

转载于:https://www.cnblogs.com/pengyu2003/p/3572425.html