面试之路（29）-合并两个排序的链表（递归和非递归）

链表的类：

class ListNode{
        int key;
        ListNode next;
    }

思路：

• 这个和数组不一样，不需要采用双指针，从后往前来

代码：

递归

public ListNode merge(ListNode head1,ListNode head2){
        if(head1 == null){
            return head2;
        }else if(head2 == null){
            return head1;
        }
        ListNode node = null;
        if(head1.key < head2.key){
            node =head1;
            node.next = merge(head1.next,head2);
        }else{
            node = head2;
            node.next = merge(head1,head2.next);
        }
        return node;
    }

非递归

public ListNode merge(ListNode head1,ListNode head2){
        if(head1 == null){
            return head2;
        }else if(head2 == null){
            return head1;
        }
        ListNode node = null;
        if(head1.key < head2.key){
            node = head1;
            head1 = head1.next;
        }else{
            node = head2;
            head2 = head2.next;
        }
        while(head1 != null && head2 != null){
            if(head1.key < head2.key){
                node.next = head1;
                head1 = head1.next;
            }else{
                node.next = head2;
                head2 = head2.next;
            }
        }
        if(head1 == null){
            node = head2;
        }
        if(head2 == null){
            node = head1;
        }
        return node;
    }