LeetCode进阶之路（Rotate List）

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Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

题目：倒着数k个node，从那开始到结尾和之前那部分对调，那个例子就是，4->5拿前面来，1->2->3拿后面去。

思路：我自己是用了三个指针，分别是移到k位置，第一位，然后把三段拼起来。思路是正确，但是太复杂，绕来绕去，后面提交也是超时了。

public ListNode rotateRight(ListNode head, int k) {ListNode result = new ListNode(-1);result.next = head;ListNode first = result;ListNode second = head;ListNode third = result;if(head == null) {return result.next;}if(head.next == null){return head;}int len = 1;while(result.next != null) {len++;result = result.next;}int i = 0;while(i < len-k) {i++;first = first.next;}third.next = first;int j = 1;while(j < len-1-k) {second = second.next;j++;}second.next = null;ListNode flag = third;while(flag.next != null) {flag = flag.next;}flag.next = head;return third.next;}

下面是参考网友的思路，用两个指针，faster/slower，先对faster设步长为n，然后faster和slower再一起走，知道faster.next==null，说明slower指向要倒着数的开始点的前一个位置。

public ListNode rotateRight(ListNode head, int n) {         if(head==null||head.next==null||n==0)            return head;        ListNode fast = head, slow = head,countlen = head;        ListNode newhead = new ListNode(-1);        int len = 0;                while(countlen!=null){            len++;            countlen = countlen.next;        }                n = n%len;        if(n==0)            return head;                for(int i = 0; i < n; i++)            fast = fast.next;                while(fast.next!=null){            slow = slow.next;            fast = fast.next;        }                newhead = slow.next;        fast.next = head;        slow.next = null;                return newhead;    }

还有一种更简单的思路，把整个链表连成一个环，在重新分割就可以了。

public ListNode rotateRight(ListNode head, int n) {    if (head == null || n == 0)        return head;    ListNode p = head;    int len = 1;//since p is already point to head    while (p.next != null) {        len++;        p = p.next;    }    p.next = head; //form a loop    n = n % len;    for (int i = 0; i < len - n; i++) {         p = p.next;    } //now p points to the prev of the new head    head = p.next;    p.next = null;    return head;}

种一棵树最好的时间是十年前，其次是现在！

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