Leetcode No.21

秒速五厘米 2022-11-19 07:50 169阅读 0赞

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L0Nsb3VkMTIwOQ_size_16_color_FFFFFF_t_70]

## JS 1° ##

我从来没用js写过链表，js好像不需要链表，好多功能可以直接用内置方法实现。有点生疏，看了题解的递归方法。但没明白这个输出，每次return的节点是怎么连起来的  
\[写完python又看了一会，终于看懂了T T\]

/** * Definition for singly-linked list. * function ListNode(val, next) { * this.val = (val===undefined ? 0 : val) * this.next = (next===undefined ? null : next) * } */
    /** * @param {ListNode} l1 * @param {ListNode} l2 * @return {ListNode} */
    var mergeTwoLists = function(l1, 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(l1, l2.next);
            return l2;
        }
    };

## python ##

# Definition for singly-linked list.
    # class ListNode:
    # def __init__(self, val=0, next=None):
    # self.val = val
    # self.next = next
    class Solution:
        def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
            if l1 is None: return l2  
            if l2 is None: return l1
            if l1.val <= l2.val:  
                l1.next = self.mergeTwoLists(l1.next,l2)
                return l1
            else:
                l2.next = self.mergeTwoLists(l1,l2.next)
                return l2

照js改的

## python 2° ##

在python题解里看到了这个方法，先把两个链表合并成有序数组，在造个假链表交回去（这就是我我看到题的最初想法）

# Definition for singly-linked list.
    # class ListNode:
    # def __init__(self, x):
    # self.val = x
    # self.next = None
    
    class Solution:
        def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
            l = []
            while l1:
                l.append(l1.val)
                l1 = l1.next
            while l2:
                l.append(l2.val)
                l2 = l2.next
            l.sort()
    
            dummy = ListNode(0)
            curr = dummy
            for i in l:
                node = ListNode(i)
                curr.next = node
                curr = curr.next
            return dummy.next

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L0Nsb3VkMTIwOQ_size_16_color_FFFFFF_t_70]: /images/20221119/245d3c673ad54a70839c09b18fece93d.png