LeeCode -- Rotate List C 语言 AC Code

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

Example 1:

image

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

Example 2:

image

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

哈哈哈哈哈哈哈 这个题目是我刷LeetCode 有史以来解决的最快的一道题目!

题目名称叫旋转链表,其实只是将尾部的结点不断的追加到链表的头部上, 如果把链表收尾相连形成一个环的话,那么各个节点的相对位置都是不发生改变的, 我们需要做的就是在K次 Rotate后 找到新的头部节点位置, 然后返回就可以了.

代码思路也很简单:

1. 获取到链表的长度和尾部的节点
2. 将链表的首尾相连, 形成链表环
3. 计算链表新的头部位置, (这里有个取余的操作, 确保 O(n) 时间内能找到新的头部位置)
4. 返回新的链表头

两次遍历, 时间复杂度为: O(n)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

typedef struct ListNode ListNode;

struct ListNode* rotateRight(struct ListNode* head, int k){
    if( !head ) return NULL;
    
    int lengthCount = 0;
    int finalStop = 0;
    ListNode *headSave = head;
    ListNode *tailPosition = NULL;
    while(head){
        lengthCount++;
        tailPosition = head;
        head = head->next;
    }
    
    head = headSave;
    finalStop = lengthCount - (k % lengthCount) - 1;
    tailPosition->next = head;
    
    while( finalStop-- ){
        head = head->next;
    }
    headSave = head->next;
    
    head->next = NULL;
    return headSave;
    
}