问题描述: 合并两个排序的链表。
解法一:用递归
解法: 递归比较结点大小:每次递归,取出两个链表的头结点来比较,比较小的结点加入新链表中。
参考代码如下
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| * 用递归 * @param head1 * @param head2 * @return */ public static ListNode merge(ListNode head1, ListNode head2) { if(head1==null) { return head2; } if(head2==null) { return head1; } ListNode head; if(head1.value<head2.value) { head=head1; head.next=merge(head1.next, head2); } else { head=head2; head.next=merge(head1, head2.next); } return head; }
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解法二:用循环
解法: 使用指针移动的方式来遍历两个链表。每次两个指针停止,进行结点比较,较小的结点加到新链表中。
参考代码如下
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| * 用循环 * @param head1 * @param head2 * @return */ public static ListNode merge1(ListNode head1, ListNode head2) { if (head1 == null) { return head2; } if (head2 == null) { return head1; } ListNode head; if (head1.value < head2.value) { head = head1; head1 = head1.next; } else { head = head2; head2 = head2.next; } ListNode move = head; while (head1 != null || head2 != null) { if (head1 == null) { move.next = head2; head2 = head2.next; } else if (head2 == null) { move.next = head1; head1 = head1.next; } else if (head1.value < head2.value) { move.next = head1; head1 = head1.next; } else { move.next = head2; head2 = head2.next; } if(move.value>move.next.value) { System.err.println("输入的链表不是有序的"); return null; } move = move.next; } return head; }
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