leetcode重排链表

题目描述：

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。

例子1：

输入：head = [1,2,3,4]
输出：[1,4,2,3]

例子2：

输入：head = [1,2,3,4,5]
输出：[1,5,2,4,3]

代码实现：

/**
 * 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:

ListNode* reverseList(ListNode* head) {
    if (head == nullptr) return nullptr;
	ListNode* pre = nullptr;
	ListNode* cur = head;
	ListNode* next1 = cur->next;
	while (next1 != nullptr) {
		cur->next = pre;
		pre = cur;
		cur = next1;
		next1 = cur->next;
	}
	cur->next = pre;
	return cur;
}

    void reorderList(ListNode* head) {
        if (head == nullptr) return;
        int listLength = 0;
        ListNode* node = head;
        while (node != nullptr) {
            listLength++;
            node = node->next;
        }
		ListNode* mid = head;
		int listLen1 = listLength / 2 + 1;
		int listLen2 = listLength - listLen1;
		for (int i = 0; i < listLen1 - 1; i++) {
			mid = mid->next;
		}
		ListNode* list2 = mid->next;
		mid->next = nullptr;
		list2 = reverseList(list2);
		ListNode* pre = head;
		ListNode* next = pre->next;
		ListNode* next2 = list2;
		while (list2 != nullptr) {
			next2 = list2->next;
			pre->next = list2;
			list2->next = next;
			pre = next;
			next = pre->next;
			list2 = next2;
		}
    }
};

标签：pre,head,ListNode,nullptr,next,链表,重排,list2,leetcode
来源： https://blog.csdn.net/Hello_Rainy/article/details/123579690