LintCode 重排链表

题目

给定一个单链表L: L0→L1→…→Ln-1→Ln,

重新排列后为：L0→Ln→L1→Ln-1→L2→Ln-2→…

必须在不改变节点值的情况下进行原地操作。

样例
给出链表 1->2->3->4->null，重新排列后为1->4->2->3->null。

分析

算法很原始，先找到链表的中间节点，然后将后半部分提出来，把后半部分反转，然后将前半部分与反转了的后半部分合并起来，合并的时候注意交叉合并就可以了

代码

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The head of linked list.
     * @return: void
     */
     private ListNode reverse(ListNode head) {
        ListNode newHead = null;
        while (head != null) {
            ListNode temp = head.next;
            head.next = newHead;
            newHead = head;
            head = temp;
        }
        return newHead;
    }

    private void merge(ListNode head1, ListNode head2) {
        int index = 0;
        ListNode dummy = new ListNode(0);
        while (head1 != null && head2 != null) {
            if (index % 2 == 0) {
                dummy.next = head1;
                head1 = head1.next;
            } else {
                dummy.next = head2;
                head2 = head2.next;
            }
            dummy = dummy.next;
            index ++;
        }
        if (head1 != null) {
            dummy.next = head1;
        } else {
            dummy.next = head2;
        }
    }

    private ListNode findMiddle(ListNode head) {
        ListNode slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    public void reorderList(ListNode head) {
        if (head == null || head.next == null) {
            return;
        }

        ListNode mid = findMiddle(head);
        ListNode tail = reverse(mid.next);
        mid.next = null;

        merge(head, tail);
    }
}