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Leetcode - 21. Merge Two Sorted Lists

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21. Merge Two Sorted Lists 解題

題目資訊

題目連結

難度: easy
使用語言: python、javascript

題目:

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.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

想法

很簡單的依照大小作排列

要注意長度不一樣長的時候，結束 while 後要補尾吧

Code

Python

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# Definition for singly-linked list.
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# class ListNode:
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#     def __init__(self, x):
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#         self.val = x
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#         self.next = None
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class Solution:
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    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
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        if not (l1 or l2):
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            return None
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        if not l1:
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            return l2
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        if not l2:
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            return l1
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        head=now = ListNode(0)
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        while l1 and l2:
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            if l1.val <= l2.val:
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                now.next = ListNode(l1.val)
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                l1 = l1.next
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            else:
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                now.next = ListNode(l2.val)
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                l2 = l2.next
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            now=now.next
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        if l1:
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            now.next = l1
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        elif l2:
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            now.next = l2
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        return head.next

python result

Javascript

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/**
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 * Definition for singly-linked list.
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 * function ListNode(val) {
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 *     this.val = val;
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 *     this.next = null;
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 * }
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 */
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/**
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 * @param {ListNode} l1
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 * @param {ListNode} l2
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 * @return {ListNode}
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 */
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var mergeTwoLists = function(l1, l2) {
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    const head = now = new ListNode(0);
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    while(l1 && l2){
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        if (l1.val <= l2.val){
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            now.next = new ListNode(l1.val);
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            l1 = l1.next;
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        }
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        else{
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            now.next = new ListNode(l2.val);
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            l2 = l2.next;
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        }
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        now = now.next
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    }
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    if(l1){
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        now.next = l1;
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    }
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    else if(l2){
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        now.next = l2;
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    }
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    return head.next
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};

javascript result