题解 | #重排链表#

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseList(ListNode *head) {
        if (!head || !head->next) {
            return head;
        }
        ListNode *res = reverseList(head->next);
        head->next->next = head;
        head->next = nullptr;
        return res;
    }

    void reorderList(ListNode *head) {
        if (!head || !head->next) {
            return;
        }
        ListNode *fast = head;
        ListNode *slow = head;
        ListNode *bef = nullptr;
        while (fast && fast->next) {
            fast = fast->next->next;
            bef = slow;
            slow = slow->next;
        }
        bef->next = nullptr;
        slow = reverseList(slow);
        ListNode *res = nullptr;
        ListNode **pt = &res;
        while (head) {
            *pt = head;
            head = head->next;
            pt = &((*pt)->next);
            *pt = slow;
            slow = slow->next;
            pt = &((*pt)->next);
        }
        if (slow) {
            *pt = slow;
            slow = slow->next;
            pt = &((*pt)->next);
        }
        head = res;
    }
};

思路：先分成两个链表，一个从前往后，一个从后往前（反转链表），然后再一个个节点合成即可。

链表结点个数可能为奇数个，所以最后反转链表可能多出来一个，特殊处理。