【LeetCode】 61. Rotate List 旋转链表(Medium)(JAVA)
题目地址: https://leetcode.com/problems/rotate-list/
题目描述:
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL
题目大意
给定一个链表,旋转链表,将链表每个节点向右移动 k 个位置,其中 k 是非负数。
解题方法
1、计算出链表的长度(因为存在:k > len),然后计算出真实需要移动位置:k % len
2、找出需要移动的上一个 node,preNode
3、断开 preNode 的 next,把所有的 node 连起来
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (head == null) return head;
ListNode temp = head;
int count = 1;
while (temp != null) {
if (temp.next == null) break;
temp = temp.next;
count++;
}
ListNode tail = temp;
k = k % count;
temp = head;
for (int i = 1; i < count - k; i++) {
temp = temp.next;
}
tail.next = head;
ListNode res = temp.next;
temp.next = null;
return res;
}
}
执行用时 : 1 ms, 在所有 Java 提交中击败了 87.45% 的用户
内存消耗 : 37.9 MB, 在所有 Java 提交中击败了 14.53% 的用户