题目

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.

解题思路

新建一个链表l3，遍历l1,l2, 每次比较l1,l2中第一个节点大小，选择较小的元素放入l3的末尾

1. 先找到l1,l2中第一个较小的元素，作为l3第一个元素
2. 遍历l1,l2，将其中元素插入l3末尾，直到l1，l2中有一个链表遍历完
3. 如果l1为空，则将l2剩余元素放到l3末尾；如果l2为空，则将l1剩余元素放到l3末尾

代码

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
    if nil == l1 {
        return l2
    }
    if nil == l2 {
        return l1
    }

    var p3 *ListNode
    p1, p2 := l1, l2
    if p1.Val < p2.Val {
        p3 = p1
        p1 = p1.Next
        p3.Next = nil
    } else {
        p3 = p2
        p2 = p2.Next
        p3.Next = nil
    }
    l3 := p3

    for ;p1 != nil && p2 != nil; {
        if p1.Val < p2.Val {
            tmp := p1
            p1 = p1.Next
            p3.Next = tmp
            p3 = p3.Next
            p3.Next = nil
        } else {
            tmp := p2
            p2 = p2.Next
            p3.Next = tmp
            p3 = p3.Next
            p3.Next = nil
        }
    }
    
    if p1 != nil {
        p3.Next = p1 
    } else {
        p3.Next = p2
    }
    return l3
}