LeetCode Solution-61

标签：一个   lin   ext   time   思路   节点   memory   next   lan   

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

思路：首先算出链表的长度（因为平移的时候以长度为周期重复），然后将链表头尾相连变成循环链表，找到头尾节点即可。

Solution:

ListNode* rotateRight(ListNode* head, int k) {
    if (!head) return head;
        
    int len = 1;
    ListNode *newHead, *tail;
    newHead = tail = head;
        
    while (tail->next) {
        tail = tail->next;
        len++;
    }
        
    tail->next = head;  //成为循环链表
        
    k %= len;
    if (k) {
        for (int i = 0; i < len - k; i++) {
            tail = tail->next;          //找到尾节点，将头结点当成第一个节点，则尾节点是第(len-k)个节点
        }
    } 
    newHead = tail->next;
    tail->next = NULL;
    return newHead;
}

性能：
Runtime: 12 ms??Memory Usage: 9.6 MB

LeetCode Solution-61

标签：一个   lin   ext   time   思路   节点   memory   next   lan   

原文地址：https://www.cnblogs.com/dysjtu1995/p/12246002.html