//给定一个单链表 L:L0→L1→…→Ln-1→Ln ,
//将其重新排列后变为: L0→Ln→L1→Ln-1→L2→Ln-2→…
//
// 你不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。
//
// 示例 1:
//
// 给定链表 1->2->3->4, 重新排列为 1->4->2->3.
//
// 示例 2:
//
// 给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.
// Related Topics 链表
// 👍 521 👎 0
/** * 思路2:对于1-2-3-4-5 * 先将链表一分为2,拿到中间节点进行反转:5-4-3 * 然后合并前后两个链表:1-2 和 5-4-3 变为1-5-2-4-3 * @param head */ public void reorderList(ListNode head) { if (head == null) { return; } ListNode fast = head; ListNode mid = head; while (fast != null) { fast = fast.next; if (fast == null) { break; } mid = mid.next; fast = fast.next; } ListNode reverseNode = reverseListNode(mid); ListNode tmpNode = new ListNode(-1); ListNode hairNode = tmpNode; int count = 0; while (head != null && reverseNode != null) { if (count % 2 == 0) { tmpNode.next = head; head = head.next; } else { tmpNode.next = reverseNode; reverseNode = reverseNode.next; } count++; tmpNode = tmpNode.next; if (head == null) { tmpNode.next = reverseNode; break; } } // 此处是细节 tmpNode.next = null; head = hairNode.next; } private ListNode reverseListNode(ListNode head) { ListNode prev = null; while (head != null) { ListNode next = head.next; head.next = prev; prev = head; head = next; } return prev; } /** * 思路一:额外内存 * 思路不难,最后一个节点处理是细节 * @param head */ public void reorderListMethod2(ListNode head) { if (head == null) { return; } List<ListNode> list = new ArrayList<>(); ListNode tmpNode = head; while (tmpNode != null) { list.add(tmpNode); tmpNode = tmpNode.next; } ListNode pNode = new ListNode(-1); ListNode result = pNode; int left = 0; int right = list.size() - 1; while (left <= right) { pNode.next = list.get(left); pNode = pNode.next; pNode.next = list.get(right); pNode = pNode.next; left++; right--; } // 这一步是关键,不加这一步后面还会带有元素。 // 比如1-2-3-4,以上之后就变成1-4-2-3-4.... pNode.next = null; head = result.next; } }