題目描述
給定一個單鏈表,把所有的奇數節點和偶數節點分別排在一起。請注意,這裏的奇數節點和偶數節點指的是節點編號的奇偶性,而不是節點的值的奇偶性。
請嘗試使用原地算法完成。你的算法的空間複雜度應爲 O(1),時間複雜度應爲 O(nodes),nodes 爲節點總數。
示例1
輸入: 1->2->3->4->5->NULL
輸出: 1->3->5->2->4->NULL
示例2
輸入: 2->1->3->5->6->4->7->NULL
輸出: 2->3->6->7->1->5->4->NULL
說明
- 應當保持奇數節點和偶數節點的相對順序。
- 鏈表的第一個節點視爲奇數節點,第二個節點視爲偶數節點,以此類推。
leecode 地址
https://leetcode-cn.com/problems/odd-even-linked-list/
代碼
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
static class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
public ListNode oddEvenList(ListNode head) {
if (head == null || head.next == null || head.next.next == null) {
return head;
}
ListNode p1 = head;
ListNode p2 = head.next;
ListNode p3 = head.next.next;
ListNode node2 = head.next;
while (true) {
p2.next = p3.next;
p2 = p2.next;
p3.next = node2;
p1.next = p3;
p1 = p1.next;
if (p2 != null && p2.next != null) {
p3 = p2.next;
} else {
break;
}
}
return head;
}
public static void main(String[] args) {
Solution solution = new Solution();
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
head.next.next.next = new ListNode(4);
head.next.next.next.next = new ListNode(5);
head.next.next.next.next.next = new ListNode(6);
head.next.next.next.next.next.next = new ListNode(7);
head.next.next.next.next.next.next.next = new ListNode(8);
// head.next.next.next.next.next.next.next.next = new ListNode(9);
print(head);
head = solution.oddEvenList(head);
print(head);
}
private static void print(ListNode head) {
while (head != null) {
System.out.print(head.val + " ");
head = head.next;
}
System.out.println();
}
}