【Leetcode】 Odd Even Linked List
2016-05-25 20:06
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题目链接:https://leetcode.com/problems/odd-even-linked-list/
题目:
Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.
You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
Example:
Given
return
Note:
The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on ...
思路:
用两个头结点 分别将奇、偶结点链接 形成两个链表 最后将两链表连接
算法:
public ListNode oddEvenList(ListNode head) {
ListNode oddHead = new ListNode(-1);
ListNode evenHead = new ListNode(-1);
ListNode p = head, op = oddHead, ep = evenHead;
int count = 0;
while (p != null) {
count++;
if (count % 2 == 1) {
op.next = p;
op = p;
} else {
ep.next = p;
ep = p;
}
p = p.next;
}
ep.next = null;
op.next = evenHead.next;
return oddHead.next;
}
题目:
Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.
You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
Example:
Given
1->2->3->4->5->NULL,
return
1->3->5->2->4->NULL.
Note:
The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on ...
思路:
用两个头结点 分别将奇、偶结点链接 形成两个链表 最后将两链表连接
算法:
public ListNode oddEvenList(ListNode head) {
ListNode oddHead = new ListNode(-1);
ListNode evenHead = new ListNode(-1);
ListNode p = head, op = oddHead, ep = evenHead;
int count = 0;
while (p != null) {
count++;
if (count % 2 == 1) {
op.next = p;
op = p;
} else {
ep.next = p;
ep = p;
}
p = p.next;
}
ep.next = null;
op.next = evenHead.next;
return oddHead.next;
}
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