• LeetCode


    Merge Two Sorted Lists

    2013.12.22 03:24

    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.

    Solution:

      This is a FAQ in IT interview. Make sure you don't new any node, and watch out for any special cases like NULL pointer or what.

      Time complexity is O(m + n), where m and n are the lengths of two lists. Space complexity is O(1).

    Accepted code:

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
    12         // IMPORTANT: Please reset any member data you declared, as
    13         // the same Solution instance will be reused for each test case.
    14         if(l1 == nullptr){
    15             return l2;
    16         }else if(l2 == nullptr){
    17             return l1;
    18         }
    19         
    20         ListNode *l3, *head;
    21         
    22         if(l1->val <= l2->val){
    23             l3 = l1;
    24             l1 = l1->next;
    25         }else{
    26             l3 = l2;
    27             l2 = l2->next;
    28         }
    29         l3->next = nullptr;
    30         head = l3;
    31         
    32         while(l1 != nullptr && l2 != nullptr){
    33             if(l1->val <= l2->val){
    34                 l3->next = l1;
    35                 l1 = l1->next;
    36             }else{
    37                 l3->next = l2;
    38                 l2 = l2->next;
    39             }
    40             l3 = l3->next;
    41             l3->next = nullptr;
    42         }
    43         
    44         while(l1 != nullptr){
    45             l3->next = l1;
    46             l1 = l1->next;
    47             l3 = l3->next;
    48             l3->next = nullptr;
    49         }
    50         while(l2 != nullptr){
    51             l3->next = l2;
    52             l2 = l2->next;
    53             l3 = l3->next;
    54             l3->next = nullptr;
    55         }
    56         
    57         return head;
    58     }
    59 };
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  • 原文地址:https://www.cnblogs.com/zhuli19901106/p/3485741.html
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