-
Notifications
You must be signed in to change notification settings - Fork 44
Commit
This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository.
refactor 2-pointer cheatsheet, add 2_pointers_linkedlist.md
- Loading branch information
Showing
2 changed files
with
176 additions
and
132 deletions.
There are no files selected for viewing
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,146 @@ | ||
| # Two pointers - Linkedlist | ||
|
|
||
| - Ref | ||
| - [fucking-algorithm : 2 pointers Linkedlist](https://labuladong.online/algo/essential-technique/linked-list-skills-summary/) | ||
|
|
||
| ### 0-1) Types | ||
|
|
||
| - Pointer types | ||
| - `Fast - Slow pointers` | ||
| - fast, slow pointers from `same start point` | ||
| - Usualy set | ||
| - slow pointer moves 1 idx | ||
| - fast pointer moves 2 idx | ||
| - linked list | ||
| - find mid point of linked list | ||
| - check if linked list is circular | ||
| - LC 141 | ||
| - LC 142 | ||
| - if a circular linked list, return beginning point of circular | ||
| - find last k elements of a single linked list | ||
| - LC 19 : Remove Nth Node From End of List | ||
|
|
||
| ## 1) General form | ||
|
|
||
| ### 1-1) Basic OP | ||
|
|
||
| #### 1-1-1 : Check if there is a circular linked list | ||
| ```java | ||
| // java | ||
| boolean hasCycle(ListNode head){ | ||
| fast = slow = head; | ||
| // NOTE : while loop condition | ||
| while (fast != null and fast.next != null){ | ||
| /** NOTE : need to do move slow, fast pointer then compare them */ | ||
| slow = slow.next; | ||
| fast = fast.next.next; | ||
| if (fast == slow){ | ||
| return True | ||
| } | ||
| } | ||
| return False; | ||
| } | ||
| ``` | ||
|
|
||
| #### 1-1-2 : return the "ring start point" of circular linked list | ||
| ```java | ||
| // java | ||
| // LC 141 | ||
| ListNode detectCycle(ListNode head){ | ||
| ListNode fast, slow; | ||
| fast = slow = head; | ||
| while (fast != null and fast.next != null){ | ||
| /** NOTE !!! We move pointers first */ | ||
| fast = fast.next.next; | ||
| slow = slow.next; | ||
| if (fast == slow){ | ||
| break; | ||
| } | ||
| } | ||
| slow = head; | ||
| // may need below logic to check whether is cycle linked list or not | ||
| // if (! fast or ! fast.next){ | ||
| // return null; | ||
| // } | ||
| while (slow != fast){ | ||
| slow = slow.next; | ||
| fast = fast.next; | ||
| } | ||
| return slow; | ||
| } | ||
| ``` | ||
|
|
||
| ```python | ||
| # LC 142. Linked List Cycle II | ||
| # python | ||
| class Solution: | ||
| def detectCycle(self, head): | ||
| if not head or not head.next: | ||
| return | ||
| slow = fast = head | ||
| while fast and fast.next: | ||
| fast = fast.next.next | ||
| slow = slow.next | ||
| if fast == slow: | ||
| break | ||
| #print ("slow = " + str(slow) + " fast = " + str(fast)) | ||
| ### NOTE : via below condition check if is a cycle linked list | ||
| if not fast or not fast.next: | ||
| return | ||
| """ | ||
| ### NOTE : re-init slow or fast as head (from starting point) | ||
| -> can init slow or head | ||
| """ | ||
| slow = head | ||
| #fast = head | ||
| """ | ||
| ### NOTE : check while slow != fast | ||
| ### NOTE : use the same speed | ||
| """ | ||
| while slow != fast: | ||
| # NOTE this !!! : fast, slow move SAME speed (in this step) | ||
| fast = fast.next | ||
| slow = slow.next | ||
| return slow | ||
|
|
||
| # V0' | ||
| # IDEA : SET | ||
| class Solution(object): | ||
| def detectCycle(self, head): | ||
| if not head or not head.next: | ||
| return | ||
| s = set() | ||
| while head: | ||
| s.add(head) | ||
| head = head.next | ||
| if head in s: | ||
| return head | ||
| return | ||
| ``` | ||
| #### 1-1-3 : find mid point of a single linked list | ||
| ```java | ||
| // java | ||
| while (fast != null and fast.next != null){ | ||
| fast = fast.next.next; | ||
| slow = slow.next; | ||
| } | ||
| return slow; | ||
| ``` | ||
|
|
||
| #### 1-1-4 : find last k elements in a single linked list | ||
| ```java | ||
| // java | ||
| ListNode fast, slow; | ||
| slow = fast = head; | ||
| while (k > 0){ | ||
| fast = fast.next; | ||
| k -= 1; | ||
| } | ||
| while (fast != null){ | ||
| fast = fast.next; | ||
| slow = slow.next; | ||
| } | ||
| return slow; | ||
| ``` | ||
|
|
||
| ## 2) LC Example |