标签:efi for rip || ini amp list tno @param
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
依次将链表最后一个元素移到最前面,共移动k次。
由于k可能大于链表本身长度length,所以先遍历一遍链表得到length,简化 k = k % length,再找到倒数k个元素前的一个元素(即新链表的最后一个元素),以此为分界点,将前半链表接到后半链表的后面。
/**
* 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 (k == 0 || head == null) {
return head;
}
int length = 1;
ListNode last = head;
while (last.next != null) {
length++;
last = last.next;
}
k = k % length;
if (k == 0) {
return head;
}
// 找到新链表的尾结点
int count = 1;
ListNode newLast = head;
while (count < length - k) {
newLast = newLast.next;
count++;
}
// 找到新链表的头结点,并将前半链表移到最后
ListNode newHead = newLast.next;
newLast.next = null;
last.next = head;
return newHead;
}
}
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} k
* @return {ListNode}
*/
var rotateRight = function (head, k) {
if (!head || !k) return head
let len = 1
let last = head
while (last.next) {
len++
last = last.next
}
k %= len
if (!k) return head
let p = head
let count = len - 1
while (count !== k) {
p = p.next
count--
}
let newHead = p.next
p.next = null
last.next = head
return newHead
}
标签:efi for rip || ini amp list tno @param
原文地址:https://www.cnblogs.com/mapoos/p/13282218.html