LeetCode-21. Merge Two Sorted Lists

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问题描述

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

Constraints:

  • The number of nodes in both lists is in the range [0, 50].
  • -100 <= Node.val <= 100
  • Both l1 and l2 are sorted in non-decreasing order.

解答

方法一：递归法
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/**
 * @Description: 合并两个有序链表
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * 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}
 * @Description: 递归法
 */
const mergeTwoLists = function (l1, l2) {
  if (l1 === null) {
    return l2
  } else if (l2 === null) {
    return l1
  } else if (l1.val < l2.val) {
    l1.next = mergeTwoLists2(l1.next, l2)
    return l1
  } else {
    l2.next = mergeTwoLists2(l1, l2.next)
    return l2
  }
}
方法二：迭代法
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/**
 * @Description: 合并两个有序链表
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * 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}
 * @Description: 迭代法
 */
const mergeTwoLists = function (l1, l2) {
  if (l1 === null) {
    return l2
  } else if (l2 === null) {
    return l1
  const listHead = new ListNode()
  let preNode = listHead

  while (l1 !== null && l2 !== null) {
    if (l1.val < l2.val) {
      preNode.next = l1
      l1 = l1.next
    } else {
      preNode.next = l2
      l2 = l2.next
    }
    preNode = preNode.next
  }
  preNode.next = (l1 === null) ? l2 : l1
  return listHead.next
}