面试之路（29）-合并两个排序的链表（递归和非递归）

来源：互联网发布：微信淘宝链接转换编辑：程序博客网时间：2024/05/22 03:45

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

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