【链表】重排链表

题目

143. 重排链表

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给定一个单链表 L 的头节点 head ，单链表 L 表示为：

L0 → L1 → … → Ln - 1 → Ln

请将其重新排列后变为：

L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1：

输入：head = [1,2,3,4]
输出：[1,4,2,3]

示例 2：

输入：head = [1,2,3,4,5]
输出：[1,5,2,4,3]

提示：

• 链表的长度范围为 [1, 5 * 104]
• 1 <= node.val <= 1000

解题

方法一：找链表中点 原地反转链表 合并链表

思路

找到链表中点后把链表分成两部分，对后面一部分链表进行原地反转，然后合并前后两部分链表。

代码

class Solution {
    public void reorderList(ListNode head) {
        if (head.next == null) {
            return;
        }

        ListNode mid = getMid(head);

        ListNode list1 = head, list2 = reverseLinkedList(mid.next);
        mid.next = null;
        mergeLinkedList(list1, list2);
    }

    public ListNode getMid(ListNode head) {
        ListNode slow = head, fast = head;
        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

    public ListNode reverseLinkedList(ListNode head) {
        ListNode curr = head, newNext = null, newPrev;
        while(curr != null) {
            newPrev = curr.next;
            curr.next = newNext;
            newNext = curr;
            curr = newPrev;
        }
        return newNext;
    }

    public void mergeLinkedList(ListNode linkedList1, ListNode linkedList2) {
        ListNode node1, node2;
        while(linkedList1 != null && linkedList2 != null) {
            node1 = linkedList1.next;
            node2 = linkedList2.next;

            linkedList1.next = linkedList2;
            linkedList1 = node1;

            linkedList2.next = linkedList1;
            linkedList2 = node2;
        }
    }
}

/**
 * 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 {
    ListNode* reverseList(ListNode* head) {
        ListNode* curr = head, * newNext = nullptr, * newPrev;
        while (curr) {
            newPrev = curr->next;
            curr->next = newNext;
            newNext = curr;
            curr = newPrev;
        }
        return newNext;
    }

    void mergeList(ListNode* list1, ListNode* list2) {
        ListNode* next1, * next2;
        while (list1 && list2) {
            next1 = list1->next;
            next2 = list2->next;
            list1->next = list2;
            list1 = next1;
            list2->next = list1;
            list2 = next2;
        }
    }

public:
    void reorderList(ListNode* head) {
        if (!head->next) return;
        ListNode* slow = head, * quick = head;
        while (quick && quick->next) {
            slow = slow->next;
            quick = quick->next->next;
        }
        ListNode* sec_half = reverseList(slow->next);
        slow->next = nullptr;
        mergeList(head, sec_half);
    }
};