来源

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

描述

将两个有序链表合并为一个新的有序链表并返回。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:
输入:1->2->4, 1->3->4
输出:1->1->2->3->4->4

题解

递归

  1. /**
  2. * Definition for singly-linked list.
  3. * public class ListNode {
  4. * int val;
  5. * ListNode next;
  6. * ListNode(int x) { val = x; }
  7. * }
  8. */
  9. class Solution {
  10. public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
  11. if (l1 == null) {
  12. return l2;
  13. } else if (l2 == null) {
  14. return l1;
  15. } else if (l1.val < l2.val) {
  16. l1.next = mergeTwoLists(l1.next, l2);
  17. return l1;
  18. } else {
  19. l2.next = mergeTwoLists(l1, l2.next);
  20. return l2;
  21. }
  22. }
  23. }

复杂度分析

  • 时间复杂度:21. 合并两个有序链表(Merge Two Sorted Lists) - 图1
  • 空间复杂度:21. 合并两个有序链表(Merge Two Sorted Lists) - 图2

迭代

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode preHead = new ListNode(-1);
        ListNode prev = preHead;

        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                prev.next = l1;
                l1 = l1.next;
            } else {
                prev.next = l2;
                l2 = l2.next;
            }
            prev = prev.next;
        }

        prev.next = (l1 == null) ? l2 : l1;

        return preHead.next;
    }
}

复杂度分析

  • 时间复杂度:21. 合并两个有序链表(Merge Two Sorted Lists) - 图3
  • 空间复杂度:21. 合并两个有序链表(Merge Two Sorted Lists) - 图4