Rotate List
Source
Problem 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
/** * 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).