[刷题]Rotate List

最新推荐文章于 2024-02-14 13:19:14 发布

原创最新推荐文章于 2024-02-14 13:19:14 发布 · 338 阅读

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本文详细介绍了如何使用O(n)复杂度实现链表的右旋转操作，并提供了关键步骤和代码实现，确保理解每一步背后的逻辑。

[LintCode]Rotate List

/**
 * 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) {
        // 2015-08-28 O(n) 
        // 这一题需要注意测试用例，k可以为0也可以很大
        if (head == null) {
            return head;
        }
        int length = getLength(head);
        k %= length;
        if (k == 0) {
            return head;
        }
        
        ListNode fast = head;
        ListNode slow = head;
        for (int i = 0; i < k; i++) {
            fast = fast.next;
        }
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        // 找到新的表头，断开并拼接链表
        ListNode firstNode = slow.next;
        slow.next = null;
        fast.next = head;
        
        return firstNode;
    }
    
    private int getLength(ListNode head) {
        int count = 0;
        while (head != null) {
            count++; 
            head = head.next;
        }
        return count;
    }
}