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
        if(head == nullptr) return head;
        vector<int> numSingle;
        vector<int> numDouble;
        ListNode* ans = new ListNode(0);
        ListNode* cur = ans;
        bool isSingle = true;
        while(head)
        {
            if(isSingle)
            {
                numSingle.push_back(head->val);
            }
            else
            {
                numDouble.push_back(head->val);
            }
            isSingle = !isSingle;
            head = head->next;
        }
        for(int val : numSingle)
        {
            ListNode* temp = new ListNode(val);
            cur->next = temp;
            cur = cur->next;
        }
        for(int val : numDouble)
        {
            ListNode* temp = new ListNode(val);
            cur->next = temp;
            cur = cur->next;
        }
        return ans->next;
    }
};