合并两个有序链表python3

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最新推荐文章于 2025-03-20 21:17:32 发布

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分类专栏： Python 文章标签：链表数据结构 leetcode

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本文详细介绍了如何实现LeetCode第21题的解决方案，即合并两个已排序的链表。首先定义了单链表节点类`classListNode`，接着通过尾插法创建了两个有序链表`l1`和`l2`。然后，通过`Solution`类的`mergeTwoLists`方法，实现了合并两个链表的功能，该方法比较链表节点的值并按顺序插入到新的链表中。最后，通过主程序展示合并过程，并打印出合并后的链表元素。

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Leetcode21题

主要实现：创建节点--创建2个单链表- 链表合并

1、创建单链表节点类

class ListNode:
    def __init__(self,val = 0, next = None):
        self.val = val
        self.next = next

2、两个有序链表创建（使用尾插法）

    l1 = [1,2,4]
    l2 = [1,3,4]
    head1 = ListNode(0)
    head2 = ListNode(0)
    cur1 = head1
    cur2 = head2
# 单链表1
    for i in range(len(l1)):
        list1 = ListNode(l1[i])
        cur1.next = list1
        cur1 = cur1.next

#单链表2
    for j in range(len(l2)):
        list2 = ListNode(l2[j])
        cur2.next = list2
        cur2 = cur2.next

3、合并这两个链表

class Solution:
    def mergeTwoLists(self,list1,list2):
        prehead = ListNode(-1)
        pre = prehead
        while list1 and list2:
            if list1.val <= list2.val:
                pre.next = list1
                list1 = list1.next
            else:
                pre.next = list2
                list2 = list2.next
            pre = pre.next
        if list1 is not None and list2 is None:
            pre.next = list1
        elif list1 is None and list2 is not None:
            pre.next = list2

        return prehead.next

4 完整程序



class ListNode:
    def __init__(self,val = 0, next = None):
        self.val = val
        self.next = next

class Solution:
    def mergeTwoLists(self,list1,list2):
        prehead = ListNode(-1)
        pre = prehead
        while list1 and list2:
            if list1.val <= list2.val:
                pre.next = list1
                list1 = list1.next
            else:
                pre.next = list2
                list2 = list2.next
            pre = pre.next
        if list1 is not None and list2 is None:
            pre.next = list1
        elif list1 is None and list2 is not None:
            pre.next = list2

        return prehead.next


if __name__ == "__main__":
    l1 = [1,2,4]
    # l1 = []
    l2 = [1,3,4]
    head1 = ListNode(0)
    head2 = ListNode(0)
    cur1 = head1
    cur2 = head2

    for i in range(len(l1)):
        list1 = ListNode(l1[i])
        cur1.next = list1
        cur1 = cur1.next


    for j in range(len(l2)):
        list2 = ListNode(l2[j])
        cur2.next = list2
        cur2 = cur2.next

    # solu = Solution()
    ans = Solution().mergeTwoLists(head1.next,head2.next)
    while ans != None:
        print(ans.val,end='')
        ans = ans.next