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Use 3 pointers each of them points to the address of Head, The node before Tail and Tail node; When rotating the list step by step, tail.next = head; tail_prev.next = null. New head = tail and new Tail = tail_prev;
Code: (Recursion way)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode rotateRight(ListNode head, int k) {
if(head == null || head.next == null) return head;
ListNode cur = head;
int len = 0;
while(cur != null && cur.next != null){
cur = cur.next;
len++;
}
if(cur != null) len++;
k = k % len;
if(k == 0) return head;
ListNode tail = cur;
ListNode newHead = rotateOneStep(head, tail, k);
return newHead;
}
public ListNode rotateOneStep(ListNode head, ListNode tail, int num){
ListNode tail_prev = head;
while(tail_prev != null && tail_prev.next.next != null) tail_prev = tail_prev.next;
tail.next = head;
tail_prev.next = null;
if(num == 1) return tail;
head = tail;
tail = tail_prev;
return rotateOneStep(head, tail, num-1);
}
}
Code(Iteration way):
public class Solution {
//ListNode newHead, newTail;
public ListNode rotateRight(ListNode head, int k) {
if(head == null || head.next == null) return head;
ListNode cur = head;
int len = 0;
while(cur != null && cur.next != null){
cur = cur.next;
len++;
}
if(cur != null) len++;
k = k % len;
ListNode tail = cur;
for(int i = 0; i < k; i++){
ListNode tail_prev = head;
while(tail_prev != null && tail_prev.next.next != null) tail_prev = tail_prev.next;
tail.next = head;
tail_prev.next = null;
head = tail;
tail = tail_prev;
}
return head;
}
Code. Use List to store each node:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode rotateRight(ListNode head, int k) {
if(head == null || head.next == null) return head;
List<ListNode> findPrev = new ArrayList<>();
ListNode cur = head;
int len = 0;
while(cur != null && cur.next != null){
findPrev.add(cur);
cur = cur.next;
len++;
}
if(cur != null) {
findPrev.add(cur);
len++;
}
k = k % len;
ListNode tail = cur;
for(int i = 0; i < k; i++){
ListNode tail_prev = findPrev.get(len-2);
tail.next = head;
tail_prev.next = null;
head = tail;
tail = tail_prev;
findPrev.add(0, findPrev.remove(len-1));
}
return head;
}
}
Jan 19 - Rotate List; Linked List; Two Pointers; Iteration & Recursion;
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原文地址：http://www.cnblogs.com/5683yue/p/5143947.html
