题解 | #合并两个排序的链表#

链表结构：

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/

解法一、常规思路

public class Solution {
    public ListNode Merge(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0), p = head; // head头结点，避免找不到链表头
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }
        p.next = l1 != null ? l1: l2;
        return head.next;
    }
}

解法二、递归解法

public class Solution {
    public ListNode Merge(ListNode l1, ListNode l2) { 
        ListNode head = new ListNode(-1); // head头结点，防止找不到头
        doMerge(head, l1, l2);
        return head.next;
    }

    // 递归辅助函数
    private void doMerge(ListNode head, ListNode l1, ListNode l2) {
        if (l1 == null || l2 == null) {
            head.next = l1 == null ? l2 : l1; // 三元运算写起来简单但是性能很差
            return ; // 递归终止条件
        }
        if (l1.val <= l2.val) {
            head.next = l1;
            doMerge(head.next, l1.next, l2);
        } else {
            head.next = l2;
            doMerge(head.next, l1, l2.next);
        }
    }
}