【链表】Leetcode 重排链表

题目讲解

143. 重排链表

算法讲解

1.使用快慢指针将链表分成两部分 2.将后半部分的链表逆置 3.使用双指针将连个链表分别连接结点在一起

/**
 * 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) {
        //找到链表的中间节点  逆置后半部分链表
        if(head == nullptr || head->next == nullptr || head->next->next == nullptr)return;
        ListNode* slow = head;
        ListNode* fast = slow->next;
        //寻找中间节点
        while(fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        //走到这里slow指向的是中间节点的前一个结点  fast指向的是后半部分链表的最后一个节点
        //逆置后半部分链表
        ListNode* newhead = new ListNode(-1);
        ListNode* cur = slow->next;
        slow->next = nullptr; //断开两个部分的链表

        //这一种头插法比较实用
        while(cur)
        {
            ListNode* NextNode = cur->next;
            cur->next = newhead->next;
            newhead->next = cur;
            cur = NextNode;
        }

        //合并链表
        ListNode* retHead = new ListNode(-1);
        ListNode* cur2 = head;
        cur = newhead->next;
        ListNode* phead = retHead;
        while(cur2)
        {
            phead->next = cur2;
            cur2 = cur2->next;
            phead = phead->next;
            if(cur)
            {
                phead->next = cur;
                cur = cur->next;
                phead = phead->next;
            }
        }
        delete retHead;
        delete newhead;
    }
};