LeetCode OJ：Reorder List（重序链表）

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}.

如上例子所示的重序链表问题，先找到中间节点，然后将链表分成两段。将第二段反转后依次插入第一段中就得到了完整的链表，代码如下所示：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void reorderList(ListNode* head) {
        if(!head || !head->next) return;
        ListNode * fastNode = head, * slowNode = head;
        while(fastNode->next){
            fastNode = fastNode->next;
            if(fastNode->next){
                slowNode = slowNode->next;
                fastNode = fastNode->next;
            }
        }
        ListNode * p1 = head;
        ListNode * p2 = slowNode->next;
        slowNode->next = NULL;//将前一段链表的最后一个节点的下一个赋为值NULL
        //将第二个节点指向的链表颠倒过来
        ListNode * prev = NULL;
        ListNode * curr = p2;
        ListNode * tmpNode = NULL;
        while(curr){
            tmpNode = curr->next;
            curr->next = prev;
            prev = curr;
            curr = tmpNode;
        }
        p2 = prev;//颠倒之后的首节点
        ListNode * tmpNode1, * tmpNode2;
        while(p2){
            tmpNode1 = p1->next;
            p1->next = p2;
            tmpNode2 = p2->next;
            p2->next = tmpNode1;
            p1 = tmpNode1;
            p2 = tmpNode2;
        }
    }
};

代码重复有点多，写的比较乱，见谅见谅。