?LeetCode刷題實戰(zhàn)61：旋轉鏈表
算法的重要性，我就不多說了吧，想去大廠，就必須要經(jīng)過基礎知識和業(yè)務邏輯面試+算法面試。所以，為了提高大家的算法能力，這個公眾號后續(xù)每天帶大家做一道算法題，題目就從LeetCode上面選 ！
今天和大家聊的問題叫做?旋轉鏈表，我們先來看題面：
https://leetcode-cn.com/problems/rotate-list/
Given a linked list, rotate the list to the right by k places, where k is non-negative.
題意
示例?1:
輸入: 1->2->3->4->5->NULL, k = 2
輸出: 4->5->1->2->3->NULL
解釋:
向右旋轉 1?步: 5->1->2->3->4->NULL
向右旋轉 2?步: 4->5->1->2->3->NULL
示例?2:
輸入: 0->1->2->NULL, k = 4
輸出: 2->0->1->NULL
解釋:
向右旋轉 1?步: 2->0->1->NULL
向右旋轉 2?步: 1->2->0->NULL
向右旋轉 3?步:?0->1->2->NULL
向右旋轉 4?步:?2->0->1->NULL
解題
public?class?Solution?{
????public?ListNode rotateRight(ListNode head, int?k) {
????????if?(head == null?|| head.next == null) {
????????????return?head;
????????}
????????int?count = 1;
????????ListNode cur = head;
????????while?(cur.next != null) {
????????????count++;
????????????cur = cur.next;
????????}
????????k = k % count;
????????if?(k == 0) {
????????????return?head;
????????}
????????cur.next = head;
????????ListNode dummy = new?ListNode(-1);
????????dummy.next = head;
????????ListNode prev = dummy;
????????for?(int?i = 0; i < count - k; i++) {
????????????prev = prev.next;
????????}
????????cur = prev.next;
????????prev.next = null;
????????return?cur;
????}
}
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