解法I 递归

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

    if l1.Val < l2.Val{
        l1.Next = mergeTwoLists(l1.Next,l2)
        return l1
    }else{
        l2.Next = mergeTwoLists(l1,l2.Next)
        return l2
    }

}

时间复杂度

O(m+n)

空间复杂度

O(m+n)

解法II 迭代法

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

	//找到小的那个结点,作为第一个结点
	var f *ListNode
	if l1.Val < l2.Val {
		f = l1
		l1 = l1.Next
	} else {
		f = l2
		l2 = l2.Next
	}

	p := f
	for l1 != nil && l2 != nil {
		if l1.Val < l2.Val {
			p.Next = l1
			l1 = l1.Next
		} else {
			p.Next = l2
			l2 = l2.Next
		}
		p = p.Next
	}
	//循环结束,如果剩下的链表还有剩余的,直接拼接
	if l1 != nil {
		p.Next = l1
	}
	if l2 != nil {
		p.Next = l2
	}

	return f
}

时间复杂度

O(m+n)

空间复杂度

O(1)