给定单链表L：L0→L1→...→Ln-1→Ln， 重新排序：L0→Ln→L1→Ln-1→L2→Ln-2→...

本题源自leetcode  143

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思路1：用快慢指针找到中节点，然后反转后半部分链表。然后俩个链表交叉插入

代码：

void reorderList(ListNode* head) {
        if (!head || !head->next) 
            return;

        // find the middle node: O(n)
        ListNode *slow = head, *fast = head;
        while (fast && fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }

        // cut from the middle and reverse the second half: O(n)
        ListNode *head2 = slow->next;
        slow->next = NULL;

        ListNode *q = head2->next;
        head2->next = NULL;
        while (q) {
            ListNode *qNext = q->next;
            q->next = head2;
            head2 = q;
            q = qNext;
        }
        ListNode* p = head;
        // merge two lists: O(n)
        for (q = head2; p; ) {
            auto t = p->next;
            p->next = q;
            p = p->next;
            q = t;
        }
        /*
        slow->next = reverse(slow->next);
        ListNode* q = slow->next; 
        //reverse(slow->next);
        while(p != slow && q ){
            ListNode * qNext = q->next;
            q->next = p->next;
            p->next = q;
            p = q->next;
            q = qNext;
        }
        */
    }

代码2：

 void reorderList(ListNode* head) {
        if (!head || !head->next) 
            return;

        // find the middle node: O(n)
        ListNode *slow = head, *fast = head;
        while (fast && fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }

        // cut from the middle and reverse the second half: O(n)
        ListNode *head2 = reverse(slow->next);
        slow->next = NULL;
        ListNode* p = head, *q = head2;
        /*
        // merge two lists: O(n)
        for (; p; ) {
            auto t = p->next;
            p->next = q;
            p = p->next;
            q = t;
        }
        */
        //reverse(slow->next);
        while(p  && q ){
            ListNode * qNext = q->next;
            q->next = p->next;
            p->next = q;
            p = q->next;
            q = qNext;
        }
    }

    ListNode* reverse(ListNode* root){
        if(!root)
            return NULL;
        ListNode* pre = NULL;
        ListNode* p = root;
        ListNode* pNext = NULL;
        while(p){
            pNext = p->next;
            p->next = pre;
            pre = p;
            p = pNext;
        }
        return pre;
    }