LeetCode 21. 合并两个有序链表

21. Merge Two Sorted Lists

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Example 1:

Input: list1 = 1,2,4, list2 = 1,3,4

Output:1,1,2,3,4,4

Example 2:

Input: list1 = \[\], list2 = \[\]

Output: \[\]

Example 3:

**Input:**list1 = \[\], list2 = 0

Output:0

Constraints:

• The number of nodes in both lists is in the range 0, 50.
• -100 <= Node.val <= 100
• Both list1 and list2 are sorted in non-decreasing order.

解法思路：

1、递归

2、迭代

法一：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        // Recursion
        // Time: O(m+n)
        // Space: O(m+n)
        if (list1 == null) {
            return list2;
        } else if (list2 == null) {
            return list1;
        } else if (list1.val < list2.val){
            list1.next = mergeTwoLists(list1.next, list2);
            return list1;
        } else {
            list2.next = mergeTwoLists(list1, list2.next);
            return list2;
        }
    }
}

法二：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        // Iterator double pointer
        // Time: O(m+n)
        // Space: O(1)
        ListNode dummyNode = new ListNode(0);
        ListNode cur = dummyNode;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                cur.next = list1;
                list1 = list1.next;
            } else {
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }
        cur.next = list1 != null ? list1 : list2;
        return dummyNode.next;
    }
}