[LeetCode]97. Reorder List链表重排序

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-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}.

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解法：仔细分析题目意思，一个简单的方法就是将后半部分链表先逆转，然后插入到前半部分。因此可以写出如下代码：

/**
 * 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 == NULL || head->next == NULL) return;
        ListNode *slow = head, *fast = head;
        while(fast->next != NULL && fast->next->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode* head1 = head;
        ListNode* head2 = slow->next;
        head2 = reverseList(head2);
        slow->next = NULL;
        while(head2 != NULL) {
            ListNode *next1 = head1->next, *next2 = head2->next;
            head1->next = head2;
            head2->next = next1;
            head1 = next1;
            head2 = next2;
        }
    }
private:
    ListNode* reverseList(ListNode* head) {
        ListNode* rHead = NULL;
        ListNode* curr = head;
        ListNode* pTail = NULL;
        while(curr != NULL) {
            ListNode* pNext = curr->next;
            if(pNext == NULL)
                rHead = curr;
            curr->next = pTail;
            pTail = curr;
            curr = pNext;
        }
        return rHead;
    }
};