[LeetCode] 61. Rotate List #61

grandyang · 2019-05-30T16:14:30Z

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Given the head of a linked list, rotate the list to the right by k places.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [4,5,1,2,3]

Example 2:

Input: head = [0,1,2], k = 4
Output: [2,0,1]

Constraints:

The number of nodes in the list is in the range [0, 500].
-100 <= Node.val <= 100
0 <= k <= 2 * 10^9

这道旋转链表的题和之前那道 Rotate Array 很类似，但是比那道要难一些，因为链表的值不能通过下表来访问，只能一个一个的走，博主刚开始拿到这题首先想到的就是用快慢指针来解，快指针先走k步，然后两个指针一起走，当快指针走到末尾时，慢指针的下一个位置是新的顺序的头结点，这样就可以旋转链表了，自信满满的写完程序，放到 OJ 上跑，以为能一次通过，结果跪在了各种特殊情况，首先一个就是当原链表为空时，直接返回NULL，还有就是当k大于链表长度和k远远大于链表长度时该如何处理，需要首先遍历一遍原链表得到链表长度n，然后k对n取余，这样k肯定小于n，就可以用上面的算法了，代码如下:

解法一：

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head) return NULL;
        int n = 0;
        ListNode *cur = head;
        while (cur) {
            ++n;
            cur = cur->next;
        }
        k %= n;
        ListNode *fast = head, *slow = head;
        for (int i = 0; i < k; ++i) {
            if (fast) fast = fast->next;
        }
        if (!fast) return head;
        while (fast->next) {
            fast = fast->next;
            slow = slow->next;
        }
        fast->next = head;
        fast = slow->next;
        slow->next = NULL;
        return fast;
    }
};

这道题还有一种解法，跟上面的方法类似，但是不用快慢指针，一个指针就够了，原理是先遍历整个链表获得链表长度n，然后此时把链表头和尾链接起来，在往后走 n - k%n 个节点就到达新链表的头结点前一个点，这时断开链表即可，代码如下:

解法二：

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head) return NULL;
        int n = 1;
        ListNode *cur = head;
        while (cur->next) {
            ++n;
            cur = cur->next;
        }
        cur->next = head;
        int m = n - k % n;
        for (int i = 0; i < m; ++i) {
            cur = cur->next;
        }
        ListNode *newhead = cur->next;
        cur->next = NULL;
        return newhead;
    }
};

Github 同步地址：

#61

类似题目：

Split Linked List in Parts

Rotate Array

参考资料：

https://leetcode.com/problems/rotate-list/

https://leetcode.com/problems/rotate-list/discuss/22715/Share-my-java-solution-with-explanation

https://leetcode.com/problems/rotate-list/discuss/22735/My-clean-C%2B%2B-code-quite-standard-(find-tail-and-reconnect-the-list)

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