Leetcode61.旋转链表
题目链接:61.旋转链表
思路:右移k个位置就是将链表右边的k个节点放到表头。那么就让1个指针从表头先走k-1步(算上表头节点此时一共经过k个节点),然后再让一个指针从表头出发,此时两个指针含有k个节点,当先走的指针到达链表尾部,那么两个指针所包含的节点就是要右移到表头的节点。
代码:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if(head == null) return head;
int len = 0;
for(ListNode p=head; p!=null; p=p.next) len++;
k = k % len;
if(k == 0) return head;
ListNode fir = head, sec = head, pre = null;
while(k-- > 1) fir = fir.next;
while(fir.next != null){
pre = sec;
fir = fir.next;
sec = sec.next;
}
fir.next = head;
pre.next = null;
return sec;
}
}
*执行用时: 1 ms
内存消耗: 37.9 MB
*