【LeetCode】Reorder List

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}.

思路：

        先利用快慢指针找到链表中点，这个中点也是重排之后的尾结点。找到中点之后将链表分为前后两个链表，再将后面的链表逆序，然后依次从两个链表中取出头结点组成新的链表

class Solution {public:    void reorderList(ListNode *head) {        if (!head || !head->next || !head->next->next) {             return;        }        // find the last and median nodes         ListNode *slow, *fast;        slow = fast = head;        while (fast != NULL) {            if (fast->next != NULL) {                slow = slow->next;                fast = fast->next->next;            } else {                break;             }        }        ListNode *prev, *curr, *succ;                prev = slow->next;        curr = prev->next;         succ = curr ? curr->next : NULL;                prev->next = NULL;        slow->next = NULL;        // reverse list        while (succ != NULL) {            curr->next = prev;            prev = curr;            curr = succ;            succ = succ->next;        }        // reverse the last two nodes        if (curr == NULL) { // there only 1 node in list.            curr = prev;        } else {             curr->next = prev;        }        ListNode *head_1 = head->next;        ListNode *head_2 = curr;        ListNode *tail = head;                // merge two lists        while (head_1 && head_2) {            tail->next = head_2; head_2 = head_2->next;            tail = tail->next;            tail->next = head_1; head_1 = head_1->next;            tail = tail->next;        }                tail->next = head_1 ? head_1 : (head_2 ? head_2 : NULL);    }};