题目:
Given a list, rotate the list to the right by k places, where k is non-negative. (Medium)
For example:
Given 1->2->3->4->5->NULL
and k = 2
,
return 4->5->1->2->3->NULL
.
分析:
利用双指针,一个指针先走k步,然后两个指针同时走,runner走到链表尾时chaser在要翻转之前位置,然后折腾一下指针指向即可。
注意:如果k 大于指针长度时,题目表述不清,应该是按照循环回来处理。所以前面需要求一下链表长度,并把k % size。
代码:
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* rotateRight(ListNode* head, int k) { 12 if (head == nullptr) { 13 return nullptr; 14 } 15 int size = 0; 16 ListNode* temp = head; 17 while (temp != nullptr) { 18 temp = temp -> next; 19 size++; 20 } 21 k = k % size; 22 ListNode* runner = head; 23 ListNode* chaser = head; 24 for (int i = 0; i < k; ++i) { 25 runner = runner -> next; 26 } 27 while (runner -> next != nullptr) { 28 runner = runner -> next; 29 chaser = chaser -> next; 30 } 31 runner -> next = head; 32 ListNode* result = chaser -> next; 33 chaser -> next = nullptr; 34 return result; 35 } 36 };