leetcode_c++：链表：Reorder List(143)

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}.

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利用快慢两个指针将链表一分为二，针对第二个子链表求倒序，最后将两个子链表合并。

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class Solution {
public:
    void reorderList(ListNode *head) {
      if(head == NULL)
        return;

        int size = 0;
        ListNode *curNode = head;
        while(curNode) {
            size++;
            curNode = curNode->next;
        }  

        int step = size - (size >> 1); // pick the larger part
        ListNode *cursor1 = head, *cursor2 = head;

      while(--step)
            cursor2 = cursor2->next;
      ListNode *tmp = cursor2;

      if(tmp != NULL) {
        cursor2 = tmp->next;
        tmp->next = NULL;
      }

        cursor2 = reverseList(cursor2);
        while(cursor2) {
            ListNode *tmp = cursor2;
            cursor2 = cursor2->next;
            tmp->next = cursor1->next;
            cursor1->next = tmp;
            cursor1 = tmp->next;
        }
    }

    ListNode* reverseList(ListNode* head) {
        if(head == NULL)
            return head;

        ListNode *curNode = head;
        while(curNode->next) {
            ListNode *nextNode = curNode->next;
            curNode->next = nextNode->next;
            nextNode->next = head;
            head = nextNode;
        }
        return head;
    }
};