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
AC代码
int calcLen(ListNode* head) {
int len = 0;
while (head) {
len++;
head = head->next;
}
return len;
}
class Solution {
public:
ListNode* rotateRight(ListNode* head, int k) {
if(!head) return NULL;
int len = calcLen(head), r = k % len;
ListNode *pre = head, *post = head;
for (int i = 0; i < r; ++i) post = post->next;
while (post->next) {
pre = pre->next;
post = post->next;
}
ListNode* thead = pre->next;
if (!thead) return head;
post->next = head;
pre->next = NULL;
return thead;
}
};
总结
求链表长度和实际旋转长度=>找到旋转后的头=>拼接原链表首尾=>切断新头节点和上一个节点的关联
