21. 合并两个有序链表

20180923-摘抄自21. 合并两个有序链表

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

解决方案

方法一：递归

/**
 * 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) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}

方法二：非递归

/**
 * 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 listNode = new ListNode(0);
        ListNode firstNode = listNode;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                listNode.next = l1;
                l1 = l1.next;
            } else {
                listNode.next = l2;
                l2 = l2.next;
            }
            listNode = listNode.next;
        }
        
        if (l1 != null) {
            listNode.next = l1;
        } else {
            listNode.next = l2;
        }
        
        return firstNode.next;
    }
}

方法三：非递归，优化

/**
 * 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) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode listNode = new ListNode(0);
        ListNode firstNode = listNode;
        while (l1 != null || l2 != null) {
            if ((l1 != null) && (l2 == null || l1.val < l2.val)) {
                listNode.next = l1;
                l1 = l1.next;
            } else {
                listNode.next = l2;
                l2 = l2.next;
            }
            listNode = listNode.next;
        }
        
        return firstNode.next;
    }
}

没什么用啊，测试时间偶然性太大了吧