题解 | #重排链表#

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/*
12345 
1， 找到中间点mid
2，链表分成两份，l1 = 12345， l2 = 34(mid->next)
mid->next = nullptr
所以：l1 = 123, l2 = 45
2,翻转l2
3, l1和l2拼接。
Q：为什么翻转l2一定会翻转自身。

*/
class Solution {
public:

    // 为什么getNum不影响，reverse会影响
    ListNode *reverse(ListNode *root) {
        if (!root) {
            return root;
        }
        ListNode *pre = nullptr;
        ListNode *curr = root;
        while(curr) {
            ListNode *next = curr->next;
            curr->next = pre;
            pre = curr;
            curr = next;
        }
        return pre;
    }
    
    ListNode *getMid (ListNode *head) {
        ListNode *slow = head, *fast = head;
        while(fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow; 
    }
     void mergeList(ListNode *l1, ListNode *l2) {
        ListNode *l1_tmp;
        ListNode *l2_tmp;
        while (l1 != nullptr && l2 != nullptr) {
            l1_tmp = l1->next;
            l2_tmp = l2->next;
 
            l1->next = l2;
            l1 = l1_tmp;
 
            l2->next = l1;
            l2 = l2_tmp;
        }
    }
    void reorderList(ListNode *head) {
        if (!head) {
            return;
        }
        
        ListNode *mid = getMid(head);
        ListNode *l1 = head;
        ListNode *l2 = mid->next;
        mid->next = nullptr;
        ListNode *l2Reverse = reverse(l2);
        mergeList(l1, l2Reverse);
    
    }
};