21. Merge Two Sorted Lists(Linked List-Easy)

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.

将两个排序的链表合并,返回一个新链表,返回的新链表也是排好序的。

解题思路:

  • 创建两个链表,一个负责保存头节点,一个负责记录比较后的结果。

Language : c

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode* newlist = (struct ListNode *)malloc(sizeof(struct ListNode));
    struct ListNode* temp = (struct ListNode *)malloc(sizeof(struct ListNode));
    newlist = temp;
    while(l1 && l2){
        if(l1->val < l2->val){
            temp->next = l1;
            l1 = l1->next;
        }
        else{
            temp->next = l2;
            l2 = l2->next;
        }
        temp = temp->next;
    }
    temp->next = l1 ? l1 : l2;
    return newlist->next;
}

Language : cpp

/**
 * 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 newlist(INT_MIN);
        ListNode *temp = &newlist;
        if(l1 == NULL && l2 == NULL){
            return NULL;
        }
        if(l1 != NULL && l2 == NULL){
            return l1;
        }
        if(l1 == NULL && l2 != NULL){
            return l2;
        }
        while(l1 && l2){
            if(l1->val < l2->val){
                temp->next = l1;
                l1 = l1->next;
            }
            else{
                temp->next = l2;
                l2 = l2->next;
            }
            temp = temp->next;
        }
        temp->next = l1 ? l1 : l2;
        return newlist.next;
    }
};

Language:python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        result = cur = ListNode(0)
        while l1 and l2:
            if l1.val < l2.val:
                cur.next = l1
                l1 = l1.next
            else:
                cur.next = l2
                l2 = l2.next
            cur = cur.next
        cur.next = l1 or l2
        return result.next

LeetCode题目汇总: https://github.com/Jack-Cherish/LeetCode

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