Let's make this one less mysterious. For 92. Reverse Linked List II, 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:
- iteration
Guide
When?
This is the kind of solution you want when the problem has structure hiding inside a messy-looking input. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are reverseBetween, reverseN.
Guide
Why?
The point of the implementation is not to make the code longer. It is to avoid doing the same thinking twice.
- 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:
- Initialize the memory or helper structure.
- Process candidates in the order the invariant expects.
- Update the answer only when the current state is valid.
- Return the value that represents the fully processed input.
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
solution.cpp
01//iteration
02//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Reverse Linked List II.
03//Memory Usage: 7.8 MB, less than 18.66% of C++ online submissions for Reverse Linked List II.
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* reverseBetween(ListNode* head, int m, int n) {
17 if(!head || !head->next) return head;
18
19 ListNode* dummy = new ListNode(-1);
20 dummy->next = head;
21 ListNode* cur = dummy;
22 ListNode* pre;
23 ListNode *rdummy = new ListNode(-1), *rtail, *rcur;
24
25 //start from dummy because we need the previous node of reversed list
26 for(int i = 0; cur; ++i){
27 // cout << i << ": " << cur->val << endl;
28 ListNode* next = cur->next;
29 if(i == m-1){
30 //the previous node of reversed list
31 pre = cur;
32 }else if(i >= m && i <= n){
33 if(i == m) rtail = cur;
34 //prepend cur to the head of reversed list
35 cur->next = rdummy->next;
36 rdummy->next = cur;
37 //connect reversed list's tail with the org list
38 if(i == n){
39 rtail->next = next;
40 break;
41 }
42 }
43 cur = next;
44 }
45
46 //connect reversed list's head with the org list
47 pre->next = rdummy->next;
48
49 return dummy->next;
50 }
51};
52
53//Approach 2: Iterative Link Reversal.
54//Runtime: 4 ms, faster than 64.33% of C++ online submissions for Reverse Linked List II.
55//Memory Usage: 7.9 MB, less than 5.08% of C++ online submissions for Reverse Linked List II.
56//time: O(N), space: O(1)
57class Solution {
58public:
59 ListNode* reverseBetween(ListNode* head, int m, int n) {
60 ListNode *dummy = new ListNode(-1);
61 dummy->next = head;
62
63 ListNode *prev = dummy, *cur = head;
64
65 //move cur "m-1" steps forward, so it become the mth node
66 int i = 1;
67 for(; i < m; ++i){
68 prev = prev->next;
69 cur = cur->next;
70 }
71
72 //rdummy: the previous node of the head of reversed list
73 ListNode *rdummy = prev;
74 //rtail: the tail of reversed list
75 ListNode *rtail = cur;
76
77 //do n-m+1 times, moving "cur" from mth node to (n+1)th node
78 //m -> (m-1), (m+1) -> m, ..., n->(n-1)
79 for(; i <= n; ++i){
80 ListNode* next = cur->next;
81 cur->next = prev;
82 prev = cur;
83 cur = next;
84 }
85
86 //prev is the tail of reversed list
87 rdummy->next = prev;
88 //cur is the next node of the tail of reversed list
89 rtail->next = cur;
90
91 return dummy->next;
92 }
93};
94
95//recursion
96//https://leetcode.com/problems/reverse-linked-list-ii/solution/242639
97//time: O(N)
98//space: O(N), space used by recursion
99//for reverse the whole list, check https://github.com/keineahnung2345/leetcode-cpp-practices/blob/master/206.%20Reverse%20Linked%20List.cpp
100class Solution {
101public:
102 ListNode* successor;
103
104 ListNode* reverseN(ListNode* head, int n){
105 if(n == 1){
106 //successor: the node after the first n nodes
107 successor = head->next;
108 return head;
109 }
110
111 ListNode* rhead = reverseN(head->next, n-1);
112
113 //head is now the new tail of reversed list
114 head->next->next = head;
115 head->next = successor;
116
117 return rhead;
118 }
119
120 ListNode* reverseBetween(ListNode* head, int m, int n){
121 if(m == 1){
122 return reverseN(head, n);
123 }
124
125 head->next = reverseBetween(head->next, m-1, n-1);
126
127 return head;
128 }
129};Cost
Complexity
Time
O(N), space: O(1)
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.