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  • (LeetCode 21)Merge Two Sorted Lists

    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.

    题目要求:

    合并两个有序链表

    注意:

    不能开辟新的结点空间

    解题思路:

    1、归并排序,创建一个新的头结点,从头到尾分别遍历两个链表,并依次比较其大小关系,每次将头指针指向小的那个。

    2、递归思想(对于为改变链表结构的题目,一般也可以采用递归的方法)

    代码:

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    
    // Merge combination method
    class Solution {
    public:
        ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
            ListNode head(0);
            ListNode *lst;
            lst=&head;
            while(l1 && l2){
                if(l1->val<=l2->val){
                    lst->next=l1;
                    l1=l1->next;
                }
                else{
                    lst->next=l2;
                    l2=l2->next;
                }
                lst=lst->next;
            }
            if(l1)
                lst->next=l1;
            if(l2)
                lst->next=l2;
            return head.next;
        }
    };
    
    // Recursive method
    class Solution {
    public:
        ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
            if(l1 == NULL) return l2;
            if(l2 == NULL) return l1;
     
            if(l1->val < l2->val) {
                l1->next = mergeTwoLists(l1->next, l2);
                return l1;
            } else {
                l2->next = mergeTwoLists(l2->next, l1);
                return l2;
            }
        }
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  • 原文地址:https://www.cnblogs.com/AndyJee/p/4461406.html
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