题解 | #链表的奇偶重排#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <vector>
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类
     * @return ListNode类
     */
    ListNode* oddEvenList(ListNode* head) {
        // write code here
        std::vector<int> arr;
        auto tmpNode = head;
        while (tmpNode != nullptr) {
            arr.push_back(tmpNode->val);
            tmpNode = tmpNode->next;
        }
        tmpNode = head;
        for (int i = 0; i < arr.size(); i = i + 2) {
            tmpNode->val = arr[i];
            tmpNode = tmpNode->next;
        }
        for (int i = 1; i < arr.size(); i = i + 2) {
            tmpNode->val = arr[i];
            tmpNode = tmpNode->next;
        }
        return head;
    }
};