Rotate List

Question

Given a list, rotate the list to the right by k places, where k is non- negative.

Example

Given 1->2->3->4->5 and k = 2, return 4->5->1->2->3.

题解

旋转链表,链表类问题通常需要找到需要处理节点处的前一个节点。因此我们只需要找到旋转节点和最后一个节点即可。需要注意的细节是 k 有可能比链表长度还要大,此时需要取模,另一个 corner case 则是链表长度和 k 等长。

Java

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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
/**
* @param head: the List
* @param k: rotate to the right k places
* @return: the list after rotation
*/
public ListNode rotateRight(ListNode head, int k) {
if (head == null) return head;
ListNode fast = head, slow = head;
int len = 1;
for (len = 1; fast.next != null && len <= k; len++) {
fast = fast.next;
}
// k mod len if k > len
if (len <= k) {
k = k % len;
fast = head;
for (int i = 0; i < k; i++) {
fast = fast.next;
}
}
// forward slow and fast
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
// return new head
fast.next = head;
head = slow.next;
slow.next = null;

return head;
}
}

源码分析

由于需要处理的是节点的前一个节点,故最终的while 循环使用fast.next != null. k 与链表等长时包含在len <= k中。

复杂度分析

时间复杂度 \[O(n)\], 空间复杂度 \[O(1)\].


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