[LintCode/LeetCode] Rotate List

Blackjun 发布于2019-08-14 15:22 / 3431人阅读

摘要：而后吾当依除取余之法，化大为小，则指针不致于越界也。后欲寻右起第结点，令快指针先行数日，及至两指针相距为，便吟鞭东指，与慢指针策马共进。快慢指针亦止于其所焉。舞动长剑，中宫直入，直取首级，而一掌劈空，已鸿飞冥冥。自此，一代天骄，霸业已成。

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.

Note

这是一道有故事的题目。
k有多大？其翼若垂天之云，则或长于链表。链表几长？北海也，天池也，其广数千里，则k亦可容于链表也。
故吾必先穷尽链表之长度，如若head为null，抑或其长可为k所整除，则此题不必再做，返回head可也。
而后吾当依除len取余之法，化大k为小k，则指针不致于越界也。后欲寻右起第k结点，令快指针先行数日，及至两指针相距为k，便吟鞭东指，与慢指针策马共进。则快指针行至null时，慢指针可至右起第k处矣。
方是时也，孤灯长夜，指飞键落。如风吹败叶，雨打窗棂。快慢指针亦止于其所焉。此时看官不妨温酒一盏，看那链表翻转，指针易位。那新头目唤作curhead，正是slow所指之人。fast舞动长剑，中宫直入，直取head首级，而slow一掌劈空，null已鸿飞冥冥。
自此，curhead一代天骄，霸业已成。

Solution

public class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        ListNode fast = head, slow = head;
        int len = 0;
        while (fast != null) {
            fast = fast.next;
            len++;
        }
        if (head == null || k % len == 0) return head;
        fast = head;
        k = k % len;
        while (k-- > 0) fast = fast.next;
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        ListNode curhead = slow.next;
        slow.next = null;
        fast.next = head;
        return curhead;
    }
}

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