143. Reorder List

题目：

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

链接： http://leetcode.com/problems/reorder-list/

6/12/2017

参考别人的思路，自己写的代码。思路，找到中间点，后半部分反转，merge这两个半个链表

注意：

1. 找到后半部分起点之后，把前半部分最后一个元素next设为null

2. 我的这个写法原始输入最中间的值会在second half的最后一个元素，所以second half length >= first half length，因此第16行要检测是否end.next是否为null

public class Solution {
    public void reorderList(ListNode head) {
        if (head == null || head.next == null || head.next.next == null) {
            return;
        }
        ListNode mid = findMid(head);
        ListNode end = reverse(mid);
        ListNode next;

        ListNode nextF, nextS;

        while (head != null && end != null) {
            nextF = head.next;
            nextS = end.next;
            head.next = end;
            if (nextF != null) {
                end.next = nextF;
            }
            head = nextF;
            end = nextS;
        }
        return;
    }
    private ListNode findMid(ListNode head) {
        ListNode slow = head, fast = head;
        ListNode prev = head;
        while (fast != null && fast.next != null) {
            fast = fast.next;
            fast = fast.next;
            prev = slow;
            slow = slow.next;
        }
        prev.next = null;
        return slow;
    }
    private ListNode reverse(ListNode head) {
        ListNode dummy = new ListNode(-1);
        ListNode next = head.next;
        while (head != null) {
            next = head.next;
            head.next = dummy.next;
            dummy.next = head;
            head = next;
        }
        return dummy.next;
    }
}

别人比较好的解法，不需要我之前那么多check

https://discuss.leetcode.com/topic/13869/java-solution-with-3-steps

更多讨论

https://discuss.leetcode.com/category/151/reorder-list