[leetCode]143. 重排链表

线性表

链表不能直接对下标进行访问，所以可以将链表的节点全部存入线性表中，然后就可以通过下标直接访问链表节点然后重建链表

class Solution {
    public void reorderList(ListNode head) {
        if (head == null) return;
        List<ListNode> list = new ArrayList<>();
        ListNode node = head;
        while (node != null) {
            list.add(node);
            node = node.next;
        }
        int i = 0, j = list.size() - 1;
        while (i < j) {
            list.get(i).next = list.get(j);
            i++;
            list.get(j).next = list.get(i);
            j--;
        }
        list.get(i).next = null;
    }
}

寻找链表中间节点+链表逆序+合并链表

可以看到目标链表是原链表右半部分反转后和左半部分合并的结果，因此解决此题可以划分为三步:

1. 找到链表的中间节点
2. 将右半部分链表反转
3. 合并左右两部分

class Solution {
    public void reorderList(ListNode head) {
        if (head == null) return;
        ListNode mid = middleNode(head);
        ListNode l1 = head;
        ListNode l2 = mid.next;
        mid.next = null;
        l2 = reverseList(l2);
        mergeList(l1, l2);
    }

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

    private ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode cur = head;
        while (cur != null) {
            ListNode tempNext = cur.next;
            cur.next = prev;
            prev = cur;
            cur = tempNext;
        }
        return prev;
    }

    private void mergeList(ListNode l1, ListNode l2) {
        while (l1 != null && l2 != null) {
            ListNode tempL1Next = l1.next;
            ListNode tempL2Next = l2.next;
            l1.next = l2;
            l1 = tempL1Next;
            l2.next = l1;
            l2 = tempL2Next;
        }
    }
}