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61. Rotate List
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
解题分析
这个环形解题法很赞！
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/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Rotate List.
* Memory Usage: 38.8 MB, less than 46.55% of Java online submissions for Rotate List.
*
* Copy from: https://leetcode-cn.com/problems/rotate-list/solution/xuan-zhuan-lian-biao-by-leetcode/[旋转链表 - 旋转链表 - 力扣（LeetCode）]
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-02-01 23:11
*/
public ListNode rotateRight(ListNode head, int k) {
if (Objects.isNull(head) || Objects.isNull(head.next) || k == 0) {
return head;
}
ListNode current = head;
int length;
for (length = 1; Objects.nonNull(current.next); length++) {
current = current.next;
}
int pass = length - k % length;
if (pass == length || pass == 0) {
return head;
}
current.next = head;
current = head;
for (int i = 0; i < pass - 1; i++) {
current = current.next;
}
ListNode result = current.next;
current.next = null;
return result;
}
/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Rotate List.
* Memory Usage: 38.7 MB, less than 55.17% of Java online submissions for Rotate List.
*/
public ListNode rotateRightLoop(ListNode head, int k) {
if (Objects.isNull(head)) {
return null;
}
if (k == 0) {
return head;
}
ListNode current = head;
int length = 0;
while (Objects.nonNull(current)) {
length++;
current = current.next;
}
int pass = length - k % length;
if (pass == length || pass == 0) {
return head;
}
current = head;
for (int i = 0; i < pass - 1; i++) {
current = current.next;
}
ListNode result = current.next;
current.next = null;
current = result;
while (Objects.nonNull(current)) {
if (Objects.isNull(current.next)) {
break;
}
current = current.next;
}
current.next = head;
return result;
}