The trick here is to name the state correctly, then let the implementation follow. For 61. Rotate List, the solution in this repository is mainly a straightforward implementation solution.
Guide
What?
We want to turn the problem statement into a smaller set of decisions the computer can repeat safely. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: straightforward implementation.
The notes already sitting in the source point us in the right direction:
- time: O(N), space: O(1)
Guide
When?
This pattern shows up when the brute force version has too many repeated checks, too many possible branches, or too much bookkeeping to do by hand. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are rotateRight.
Guide
Why?
The solution works because it narrows the problem until every update has a clear reason to exist.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Read the setup variables first.
- Follow the main loop or recursive helper next.
- Watch where invalid states get skipped.
- Check which value survives to the return statement.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(N), space: O(1)
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
01//Runtime: 8 ms, faster than 81.03% of C++ online submissions for Rotate List.
02//Memory Usage: 11.9 MB, less than 20.29% of C++ online submissions for Rotate List.
03//time: O(N), space: O(1)
04/**
05 * Definition for singly-linked list.
06 * struct ListNode {
07 * int val;
08 * ListNode *next;
09 * ListNode() : val(0), next(nullptr) {}
10 * ListNode(int x) : val(x), next(nullptr) {}
11 * ListNode(int x, ListNode *next) : val(x), next(next) {}
12 * };
13 */
14class Solution {
15public:
16 ListNode* rotateRight(ListNode* head, int k) {
17 ListNode *slow = head, *fast = head;
18 int len = 0;
19
20 for(ListNode* cur = head; cur; cur = cur->next) ++len;
21
22 if(len <= 1) return head;
23
24 k = k % len;
25 if(k == 0) return head;
26
27 for(int i = 0; i < k; ++i) fast = fast->next;
28
29 while(slow->next && fast->next){
30 slow = slow->next;
31 fast = fast->next;
32 }
33
34 //now slow is the previous node of the new head
35 ListNode* newhead = slow->next;
36 slow->next = nullptr;
37 //fast is the last node of original list
38 fast->next = head;
39
40 return newhead;
41 }
42};
43
44//cleaner
45//https://leetcode.com/problems/rotate-list/discuss/22735/My-clean-C%2B%2B-code-quite-standard-(find-tail-and-reconnect-the-list)
46//Runtime: 8 ms, faster than 81.03% of C++ online submissions for Rotate List.
47//Memory Usage: 12 MB, less than 14.57% of C++ online submissions for Rotate List.
48class Solution {
49public:
50 ListNode* rotateRight(ListNode* head, int k) {
51 if(!head || !head->next || k == 0) return head;
52 ListNode* cur = head;
53 int len = 1;
54
55 //find tail
56 while(cur->next){
57 ++len;
58 cur = cur->next;
59 }
60 //now cur is the last node of original list
61
62 //connect tail and head
63 cur->next = head;
64
65 //find the tail of new list
66 for(int i = 0; i < len - (k%len); ++i){
67 cur = cur->next;
68 }
69
70 //head of new list is the next node of tail of new list
71 head = cur->next;
72 cur->next = nullptr;
73
74 return head;
75 }
76};Cost