题解 | #链表的奇偶重排#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
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) {
        ListNode *p = head;
    	ListNode *slow, *fast = nullptr;
    	int count = 1;
    	if (head != nullptr) {
        	if (head->next != nullptr && head->next->next != nullptr) {
            	fast = head->next->next;
            	slow = head->next;
            	while (fast != nullptr) {
                	if (count % 2 != 0) {
                    	slow->next = fast->next;
                    	fast->next=p->next;
                    	p->next = fast;
                    	p=fast;
                    	fast = slow->next;
                	} else {
                    	slow = slow->next;
                    	fast = fast->next;
                	}
                	count++;
            	}
        	}
        // write code here
        }
        return head;
    }
};

c++快慢指针，p来存储偶数下标的前一个结点，每当count为奇数次时把快指针指的值插入到p结点后面；count为偶数此时，快慢指针往后移动，直到快指针为空时停止。