Leetcode - 61. Rotate List

Given the head of a linked list, rotate the list to the right by k places. Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [4,5,1,2,3]

Example 2:

Input: head = [0,1,2], k = 4
Output: [2,0,1]

Constraints:

• The number of nodes in the list is in the range [0, 500].
• 100 <= Node.val <= 100
• 0 <= k <= 2 * 109

Java

class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if(head == null || head.next == null) return head; 
        // 创建一个指针
        ListNode pointer = head; 
        
        // 获取list size
        int size = 1; 
        while(pointer.next != null) {
            size++; 
            pointer = pointer.next; 
        }; 
        
        // 获取实际rotate的次数
        k %= size; 
        if(k == 0) return head; 
        
        // 此时，指针已处于list末尾，将末尾的node与第一个node连接起来
        pointer.next = head; 
        // 将指针重新指向list原来的第一个node
        pointer = head; 
        
        // 让指针指向rotate后的最后一个node
        for(int i = 1; i < size - k; i++) {
            pointer = pointer.next; 
        }
            
        // 新的第一个node就是指针的下一个node
        head = pointer.next; 
        // 将最后一个node的next指向null
        pointer.next = null; 
        
        return head; 
    }
}

JavaScript

var rotateRight = function(head, k) {
    
    if (head === null || head.next === null) return head;
    let pointer = head;

    // find the size of the list
    let size = 1;
    while (pointer.next !== null) {
        size++; 
        pointer = pointer.next; 
    }
        
    k %= size; 
    if(k === 0) return head; 
    
    pointer.next = head; 
    pointer = head; 
    for(let i = 1; i < size - k; i++)
        pointer = pointer.next; 
    
    head = pointer.next; 
    pointer.next = null; 
    
    return head;
};

总结

1. 题目本身难度是medium，不算难；
2. 关键点还是要先画图，认真分析，有的比较完善的思路之后再来写代码。