61. Rotate List

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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