0021-leetcode算法实现之合并两个有序链表-merge-two-sorted-lists-python&golang

将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例 1：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]
示例 2：

输入：l1 = [], l2 = []
输出：[]
示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

提示：

两个链表的节点数目范围是 [0, 50]
-100 <= Node.val <= 100
l1 和 l2 均按 非递减顺序 排列

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/merge-two-sorted-lists

python

# 0021.合并两个有序链表
# https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/xin-shou-you-hao-xue-hui-tao-lu-bu-fan-cuo-4nian-l/
class ListNode:
    def __init__(self, val):
        self.val = val
        self.next = None

class Solution:
    def mergeTwoLists1(self, l1: ListNode, l2: ListNode) -> ListNode:
        """
        递归法
        :param l1:
        :param l2:
        :return:
        """
        # 递归终止条件
        if not l1: return l2
        if not l2: return l1
        # 选第一个节点值小的作为最后返回的链表头结点
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists1(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists1(l1, l2.next)
            return l2


    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        """
        迭代法，虚拟头结点， 两个链表各遍历一遍，时间O(n),空间O(1)
        :param headA:
        :param headB:
        :return:
        """
        dummy = ListNode(0)
        # 遍历游标
        tmp = dummy
        # l1 & l2 都未遍历
        while l1 and l2:
            # 如果l1节点值小
            if l1.val < l2.val:
                # l1节点作为tmp的下个节点
                tmp.next = l1
                # l1指向下个节点
                l1 = l1.next
            else:
                # l2节点值小，l2作为tmp的下个节点
                tmp.next = l2
                # l2指向下个节点
                l2 = l2.next
            # 移动结果链表的结尾指针
            tmp = tmp.next
        # 非空节点作为最后节点
        tmp.next = l1 or l2
        # 返回合并链表的头结点
        return dummy.next

golang

// 迭代
func mergeTwoLists1(l1, l2 *ListNode) *ListNode {
	dummy := &ListNode{}
	tmp := dummy
	for l1 != nil && l2 != nil {
		if l1.Val < l2.Val {
			tmp.Next = l1
			l1 = l1.Next
		} else {
			tmp.Next = l2
			l2 = l2.Next
		}
		tmp = tmp.Next
	}
	if l1 != nil {
		tmp.Next = l1
	} else {
		tmp.Next = l2
	}
	return dummy.Next
}

// 递归
func mergeTwoLists(l1, l2 *ListNode) *ListNode {
	if l1 == nil {
		return l2
	}
	if l2 == nil {
		return l1
	}
	if l1.Val < l2.Val {
		l1.Next = mergeTwoLists(l1.Next, l2)
		return l1
	} else {
		l2.Next = mergeTwoLists(l1, l2.Next)
		return l2
	}
}