Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2 Output: 4->5->1->2->3->NULL Explanation: rotate 1 steps to the right: 5->1->2->3->4->NULL rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4 Output: 2->0->1->NULL Explanation: rotate 1 steps to the right: 2->0->1->NULL rotate 2 steps to the right: 1->2->0->NULL rotate 3 steps to the right: 0->1->2->NULL rotate 4 steps to the right: 2->0->1->NULL
难度:medium
题目:给定链表,向右旋转k个位置,k为非负整数。
思路:首先统计结点个数,并记录尾结点。然后用前后指针找出倒数第k + 1个结点。
Runtime: 7 ms, faster than 92.47% of Java online submissions for Rotate List.
Memory Usage: 27 MB, less than 41.82% of Java online submissions for Rotate List.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode rotateRight(ListNode head, int k) { if (null == head || k <= 0) { return head; } int cnt = 0; ListNode ptr = head,tail = ptr; while (ptr != null) { tail = ptr; ptr = ptr.next; cnt++; } k = k % cnt; ListNode dummyHead = new ListNode(0), lastKPtr = dummyHead; dummyHead.next = head; ptr = dummyHead.next; while (ptr != null) { if (--k < 0) { lastKPtr = lastKPtr.next; } ptr = ptr.next; } tail.next = dummyHead.next; dummyHead.next = lastKPtr.next; lastKPtr.next = null; return dummyHead.next; } }
原文
https://segmentfault.com/a/1190000018121671
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