[LeetCode] 21. Merge Two Sorted Lists ☆

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

解法：

　　新建一个链表，依次比较两个链表的头元素，把较小的移到新链表中，直到有一个为空，再将另一个链表剩余元素移到新链表末尾。

采用循环的方式，代码如下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode res = new ListNode(0);
        ListNode last = res;
        
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                last.next = l1;
                l1 = l1.next;
            } else {
                last.next = l2;
                l2 = l2.next;
            }
            last = last.next;
        }
        
        last.next = (l1 != null) ? l1 : l2;
        return res.next;
    }
}

采用递归的方式，代码如下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode head = l1.val < l2.val ? l1 : l2;
        ListNode other = l1.val < l2.val ? l2 : l1;
        head.next = mergeTwoLists(head.next, other);
        return head;
    }
}

或者：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public 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;
        }
    }
}