61. Rotate List
problem description
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
algorithm thought
可以看到其实就是将链表倒数第k个位置截断,然后拼接到前面。知道这个意思之后,就只需要考虑如果k大于size的问题了。如果k等于size,链表长度不变,这就很明显,我们可以令k=k%size。最后截断链表即可
code
/**
* 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 size=1;
ListNode* tail=head;
while(tail->next){
size++;
tail=tail->next;
}
if(size==1)
return head;
k=k%size;
if(k==0)
return head;
k=size-k-1;
ListNode* pre=head;
ListNode* cur=head->next;
while(k--){
pre=pre->next;
cur=cur->next;
}
pre->next=NULL;
tail->next=head;
return cur;
}
};
algorithm analysis
开始算法得到链表的size,时间复杂度O(n)。然后找到倒数第k个节点,最后
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