【LeetCode】61. Rotate List
2017-03-09 17:51
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题目描述
Given a list, rotate the list to the right by k places, where k is non-negative.For example:
Given
1->2->3->4->5->NULLand
k = 2,
return
4->5->1->2->3->NULL.
解题思路
先遍历一遍确定链表中元素的个数n,然后k对n求模,因为移动n的倍数还是链表本身。然后找到倒数第k个,这个就是移动之后的新的链表头。
AC代码
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* rotateRight(ListNode* head, int k) { if (head == NULL) return head; int n = 1; ListNode *tail = head; while (tail->next != NULL) { n += 1; tail = tail->next; } k = k % n; if (k == 0) return head; // 将当前链表的尾的下一个元素指向开头 tail->next = head; //找到倒数第k个,作为移动后的头元素,它的前一个就是队尾,设其next为null for (int i = 1; i < n- k; ++i) { head = head->next; } ListNode* ans = head->next; head->next = NULL; return ans; } };
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