题解 | #链表的奇偶重排#

链表的奇偶重排

https://www.nowcoder.com/practice/02bf49ea45cd486daa031614f9bd6fc3

import java.util.*;

public class Solution {
    public ListNode oddEvenList (ListNode head) {
        if(head == null || head.next == null)
            return head;
        List<Integer> yuan = new ArrayList<>();
        ListNode result = new ListNode(1);
        ListNode curr = result;
        while(head != null) {
            yuan.add(head.val);
            head = head.next;
        }
        for (int i = 0; i < yuan.size(); i = i + 2) {
            curr.next = new ListNode(yuan.get(i));
            curr = curr.next;
        }
        for (int i = 1; i < yuan.size(); i = i + 2) {
            curr.next = new ListNode(yuan.get(i));
            curr = curr.next;
        }
        return result.next;
    }
}

全部评论

相关推荐

点赞 收藏 评论
分享
牛客网
牛客企业服务