力扣刷题笔记：143. 重排链表

题目：143. 重排链表

算法：

• 快慢指针寻找中间值
• 翻转后半部分的链表
• 两部分链表相加

源代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
    
    
public:
    void reorderList(ListNode* head) {
    
    
        if(head == nullptr){
    
    
            return;
        }
        if(head->next == nullptr){
    
    
            return ;
        }
        if(head->next->next == nullptr){
    
    
            return ;
        }
        if(head->next->next->next == nullptr){
    
    
            head->next->next->next = head->next;
            head->next = head->next->next;
            head->next->next->next = nullptr;
            return;
        }
        ListNode* save_head = head;
        ListNode* fast = head->next;
        ListNode* slow = head->next;

        //寻找中间结点
        while(fast != nullptr && fast->next != nullptr){
    
    
            fast = fast->next->next;
            slow = slow->next;

            if(fast ->next != nullptr){
    
    
                //偶数个的时候，指向最后一个结点
                if(fast->next->next == nullptr){
    
    
                    fast = fast->next;
                }
            }   
        }
        //翻转后半部分链表
        ListNode* pre = nullptr;
        ListNode* curr = slow->next;
        while(curr != nullptr){
    
    
            ListNode* t = curr->next;
            curr->next = pre;
            pre = curr;
            curr = t;
        }
        slow->next = nullptr;
        
        //链表合并
        ListNode* temp = nullptr;
        while(fast != nullptr){
    
    
            temp = fast->next;
            fast->next = head->next;
            head->next = fast;
            fast = temp;
            head = head->next->next;

        }
        head = save_head;
    }
};