重排链表

合集 - 每日一练(7)

1.打印沙漏2023-07-082.栈->队列2023-07-103.反转链表2023-07-114.合并两个排序的链表2023-07-135.链表中环的入口结点2023-07-146.二叉树典型例题2023-07-18

7.重排链表2023-09-11

收起

 解题思路：采用快慢指针的方法将单链表分成两半，再对两部分的链表进行合并

void reorderList(struct ListNode* head) {

    if (head == NULL || head->next == NULL) {

        return;

    }

   

    // 找到链表中间节点

    struct ListNode* slow = head;

    struct ListNode* fast = head;

   

    while (fast != NULL && fast->next != NULL) {

        slow = slow->next;

        fast = fast->next->next;

    }

   

    // 翻转后半部分链表

    struct ListNode* prev = NULL;

    struct ListNode* curr = slow;

    struct ListNode* nextTemp;

   

    while (curr != NULL) {

        nextTemp = curr->next;

        curr->next = prev;

        prev = curr;

        curr = nextTemp;

    }

   

    // 合并两个链表

    struct ListNode* firstHalf = head;

    struct ListNode* secondHalf = prev;

   

    while (secondHalf->next != NULL) {

        struct ListNode* firstNextTemp = firstHalf->next;

        struct ListNode* secondNextTemp = secondHalf->next;

        firstHalf->next = secondHalf;

        secondHalf->next = firstNextTemp;

        firstHalf = firstNextTemp;

        secondHalf = secondNextTemp;

    }

}