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leetcode-21-merge two sorted lists

admin 11月 12, 2020 0

题目

合并两个排好续的链表

分析

略

C++代码实现



 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* (ListNode* l1, ListNode* l2) {
        ListNode tmp = ListNode(-1);
        ListNode* res = &tmp;
        while(l1 && l2)
        {
            if(l1->val < l2->val){
                res->next = l1;
                l1 = l1->next;
            }
            else{
                res->next = l2;
                l2 = l2->next;               
            }
            res = res->next;
        }
        res->next = l1 ? l1:l2;
        return tmp.next;
    }
};



 /*
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode tmp = ListNode(-1);
        ListNode* res = &tmp;
        while(l1 && l2)
        {
            if(l1->val < l2->val){
                res->next = l1;
                l1 = l1->next;
                res = res->next;
            }
            else{
                res->next = l2;
                l2 = l2->next;
                res = res->next;
            }
        }
        if(!l1){
            while(l2){
                res->next = l2;
                l2 = l2->next;
                res = res->next;
            }
        }
        else{
            while(l1){
                res->next = l1;
                l1 = l1->next;
                res = res->next;
            }
        }
        return tmp.next;
    }
};
*/

Python代码实现


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

class Solution(object):
    def (self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        res = ListNode(0)
        tmp = res
        while l1 is not None and l2 is not None:
            if l1.val < l2.val:
                tmp.next = ListNode(l1.val)
                l1 = l1.next
                tmp = tmp.next
            else:
                tmp.next = ListNode(l2.val)
                l2 = l2.next
                tmp = tmp.next
        if l1 is None:
            tmp.next = l2
        elif l2 is None:
            tmp.next = l1
        return res.next