LeetCode之Reorder List

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 /*
 1.先找到链表的中点；
 2.将链表从中间断开，然后翻转后半段链表；
 3.用尾插法将后半段链表插入前半段的链表中（相隔一个节点再插入）。
 */
class Solution {
public:
    void reorderList(ListNode *head) {
        if(head == nullptr || head->next == nullptr) return;
        
        //获得链表的后半段
        ListNode *pre(head), *slow(head), *fast(head);
        while(fast != nullptr && fast->next != nullptr){
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode *q = slow;
        q = reverse(q);//翻转后半部分的链表

        pre->next = nullptr;
        ListNode *p = head;
        ListNode *tmp;
        while(p != nullptr){//尾插法插入后半部分的链表结点
            tmp = q;
            q = q->next;
            tmp->next = p->next;
            p->next = tmp;
            p = tmp->next;//下一个插入位置
        }
        if(q != nullptr) tmp->next = q;
    }
    
    ListNode* reverse(ListNode *head){
        if(head == nullptr || head->next == nullptr) return head;
        ListNode dummy(-1);
        dummy.next = head;
        ListNode *p(head->next);
        head->next = nullptr;
        ListNode *tmp;
        while(p != nullptr){//头插法，翻转链表
            tmp = p;
            p = p->next;
            tmp->next = dummy.next;
            dummy.next = tmp;
        }
        return dummy.next;
    }
};