Leetcode: Rotate List

最新推荐文章于 2017-02-28 15:27:00 发布

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本文介绍了如何将链表向右旋转k个位置，并详细解释了实现步骤和代码逻辑。

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Problem:

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

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

Solution: Just Note that should count the length first, and rotate the list to the right by n % length places.

Code:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int n) {
        if (head == null) return head;
        int length = 0;
        ListNode forCount = head;
        ListNode newHead = new ListNode(0);
        ListNode last = newHead;
        newHead.next = head;
        while(forCount != null) {
            length++;
            forCount = forCount.next;
            last = last.next;
        }
        int k = n % length;
        if (k == 0 || (head.next == null && k == 1)) return head;
        int step= length - k;
        
        ListNode current = newHead;
        for (int i = 0; i < step; i++) {
            current = current.next;
        }
        ListNode next = current.next;
        current.next = null;
        last.next = newHead.next;
        return next;
        
    }
}