题解 | #链表的奇偶重排#

双指针
常规题

/**
 * 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) {
        // write code here
        if(head == NULL) return NULL;
        ListNode* odd = new ListNode(-1);
        ListNode* even = new ListNode(-1);
        ListNode* p = odd;
        ListNode* q = even;
        ListNode* cur = head;
        int cnt = 1;
        while(cur != NULL){
            if(cnt % 2 == 1){
                p -> next = cur;
                p = p -> next;
            }else{
                q -> next = cur;
                q = q -> next;
            }
            cur = cur -> next;
            cnt ++;
        }
        p -> next = q -> next = NULL;
        p -> next = even -> next;
        return odd -> next;
    }
};