Leetcode 61.旋转链表.md

Leetcode 61. 旋转链表

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。

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思路：

1. 计算链表长度
2. 找到 k取模链表长度 (k % length) 开始的子链表
3. 拼接子链表和原链表(被截取后的部分)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if(k == 0 || head == nullptr || head->next == nullptr) {
            return head;
        }
        int len = 0;
        ListNode* p1 = head;
        ListNode* tail = head;
        while(p1) {
            ++len;
            tail = p1;
            p1 = p1->next;
        }
        k = k % len;

        ListNode* pre = findFromEnd(head, k + 1);//倒数第K+1
        tail->next = head;
        head = pre->next;
        pre->next = nullptr;

        return head;
    }

    ListNode* findFromEnd(ListNode* head, int N) {
        ListNode* fast = head;
        ListNode* slow = head;
        while(N--) {
            fast = fast->next;
        }
        while(fast) {
            fast = fast->next;
            slow = slow->next;
        }
        return slow;
    }
};