<aside> 📚 Key Points:

When dereferencing a ListNode, make sure it is not a NULL pointer.
Never ever lose control of the head/newHead pointer of a Llinked List
A ListNode head reference is the only way to represent the entire Linked List.
Good Coding Practices:
- Better not to move head directly; create a cur pointer to traverse the list instead
- Coner case: head == null || head.next == null || k < 0 </aside>

易错点/Take-away Summary

[ ] 每次过例子，准备一个全面的例子（cover all cases），把例子完整的过完，判断需要什么指针，指针怎么移动，有什么cases，终止条件是什么。不要过完前一两个，发现规律后，马上写代码。
[ ] 改变箭头方向之后，需要断开的连接“一定要断开”。例如，找中点题里，mid.next = null; dedup题里，slow.next = null
[ ] output的头节点会改变？ —> Use dummy node to secure the new head
[ ] use dummy node to cover edge cases to make the code-writing easy and clean
[ ] dummy node 一定要跟 head连接在一起
[ ] 用dummy node方便的把list group到需要的小组，最后再拼接。group之后再拼接时，连接新的group link之后一定要把旧的link断开。例如，partition list 题，和 even odd 题
[ ] When dereferencing a ListNode, make sure it is not a NULL pointer.
[ ] A ListNode head reference is the only way to represent the entire Linked List.
[ ] Better not to move head directly; create a cur pointer to traverse the list instead
[ ] 每一轮的traversal，一定要移动所有的applicable指针。而且，顺序matters 顺序matters 顺序matters。错误的指针移动顺序可能导致丢失下一个节点的control
[ ] 移动指针时，其中一个list变成null怎么办？所有的corner case都要照顾到位 (corner case疯狂多的题： Add Two Numbers)
[ ] 写 while loop 条件或者 if conditional 的时候，一定到确保 condition里面的值不会有NPE。in while loop, must write”cur ≠ null” before writing “cur.value < target” and move cur pointer within the loop.
[ ] Cycle exists in a Linked List == ListNode address can be seen twice durign traversal
[ ] 两两比较完，post processing 别忘了 - 把剩下的/没比的node全部接到最后面。例如： merge two sorted array
[ ] When i create a new/next pointer to store the next node of cur node and change the arrow direction, ensure I call the nextNode as “next not cur.next”
[ ] NPE NPE NPE - The biggest issue from Linked List problem. For any if conditionals, make sure to “consider the null case first before considering any other non-null cases.”
[ ] Make sure “not to lose control of head & newHead”. If needed, create a dummyNode to keep the code clear and clean.
[ ] 每个step千万不要忘记移动指针
[ ] 过例子的时候，想好要移动多少个步数，count nodes between the range to see how many steps I need to move. for(int i = 0; i < numOfNodesINeedToMove; i++) (i.e LeetCode 92)
leftNode定位时，for loop 要移动 left - 1个node
rightNode定位时，for loop 需要移动 right - left + 1个node
[ ] 每次for loop完/reverse每个node之后，指针已经移动到下一轮要遍历的node上。所有出了for loop之后，要清楚的知道你的指针到底在哪里。

<aside> 🌵 Table of Contents

</aside>

Two Pointers

<aside> 📌 Two Pointers Problem Summary**:**

output的头节点会改变？ —> Use dummy node to secure the new head
每一轮的traversal，一定要移动所有的applicable指针。而且，顺序matters 顺序matters 顺序matters。错误的指针移动顺序可能导致丢失下一个节点的control
移动指针时，其中一个list变成null怎么办？所有的corner case都要照顾到位(i.e. Add Two Numbers)
写 while loop 条件，或者 if conditional 的时候，一定到确保 condition里面的值不会有NPE。in while loop, must write”cur ≠ null” before writing “cur.value < target” and move cur pointer within the loop. </aside>

Date: June 15, 2023

Date: June 16, 2023

Date: June 19, 2023

Problem 2: Add Two Numbers

Date: June 24, 2023