[LeetCode]Reorder List

题目描述：

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}.

解题方案：

类似于将两个链表合成一个链表。1.首先利用快慢指针找到链表的中点，将链表分为两段；2.反转后一链表，合并两个链表。下面是该题代码：

class Solution {
public:
    void reorderList(ListNode *head) {
        
        if (head == NULL || head->next == NULL || head->next->next == NULL) {return;}
        struct ListNode *fast = head;
        struct ListNode *slow = head;

         while(fast != NULL && fast->next != NULL) {
            fast = fast->next->next;
            slow = slow->next;
        }

        //cout <<slow->val;
        struct ListNode *CurPos = slow->next;
        slow->next = NULL;

        //cout << CurPos->val;
        //cout <<fast->val;
        //反转链表
        struct ListNode *PreCurPos = NULL;
        struct ListNode *NextCurPos = CurPos->next;

        while(CurPos != NULL) {
            CurPos->next = PreCurPos;
            //cout<<CurPos->val<<endl;
            PreCurPos = CurPos;
            if (NextCurPos == NULL) {
                break;
            }
            CurPos = NextCurPos;
            NextCurPos = CurPos->next;
        }
        slow = head;
        head = merge(slow, CurPos);
    }

    ListNode *merge(ListNode *lhs, ListNode *rhs) {
        ListNode *newhead = new ListNode(0);
        ListNode *p = newhead;
        int flag = 0;
        while (lhs && rhs) {
            if (flag == 0) {
                p->next = lhs;
                lhs = lhs->next;
                flag = 1;
            } else {
                p->next = rhs;
                rhs = rhs->next;
                flag = 0;
            }
            p = p->next;
        }
        if (lhs == NULL) {
            p->next = rhs;
        } else {
            p->next = lhs;
        }
        p = newhead->next ;
        newhead->next = NULL;
        delete newhead;
        return p;
    }
};