[LeetCode] 21. Merge Two Sorted Lists（合并两个有序链表）

• Difficulty: Easy

• Related Topics: Linked List

• Link: https://leetcode.com/problems/merge-two-sorted-lists/

Description

Merge two sorted linked lists and return it as a new sorted list. The new list should be made by splicing together the nodes of the first two lists.

合并两个有序的链表，返回合并后的链表。新链表应该是通过重组前面链表的节点构成。

Examples

Example 1

Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2

Input: l1 = [], l2 = []
Output: []

Example 3

Input: l1 = [], l2 = [0]
Output: [0]

Constraints

1. The number of nodes in both lists is in the range [0, 50].

   两个链表的节点数量在 [0, 50] 之间

2. -100 <= Node.val <= 100

3. Both l1 and l2 are sorted in non-decreasing order.

   两个链表以非递减顺序排序

Solution

合并单链表也是数据结构课的基础知识了，具体实现过程不再赘述。先给出复制节点值的方法（当然，此法不符合题目要求，不过对于我而言，这种方法更直观一点），代码如下：

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun mergeTwoLists(l1: ListNode?, l2: ListNode?): ListNode? {
        if (l1 == null && l2 == null) {
            return null
        }
        val dummyHead = ListNode(-1)
        var i: ListNode? = dummyHead
        var p = l1
        var q = l2

        while (!(p == null && q == null)) {
            val leftVal = p?.`val`?:Int.MAX_VALUE
            val rightVal = q?.`val`?:Int.MAX_VALUE

            if (leftVal < rightVal) {
                i?.next = ListNode(leftVal)
                i = i?.next
                p = p?.next
            } else {
                i?.next = ListNode(rightVal)
                i = i?.next
                q = q?.next
            }
        }

        val result = dummyHead.next
        dummyHead.next = null
        return result
    }
}

然后再给一种直接对节点进行操作的方法（来自于我看的一本算法书，原始代码为 Java 实现，我改为用 Kotlin 实现）：

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun mergeTwoLists(l1: ListNode?, l2: ListNode?): ListNode? {
        if (l1 == null || l2 == null) {
            return l1 ?: l2
        }
        val head = if (l1.`val` < l2.`val`) l1 else l2
        var cur1: ListNode? = if (head === l1) l1 else l2
        var cur2: ListNode? = if (head === l1) l2 else l1
        var pre: ListNode? = null

        while (cur1 != null && cur2 != null) {
            if (cur1.`val` <= cur2.`val`) {
                pre = cur1
                cur1 = cur1.next
            } else {
                val next = cur2.next
                pre?.next = cur2
                cur2.next = cur1
                pre = cur2
                cur2 = next
            }
        }
        pre?.next = cur1 ?: cur2
        return head
    }
}