234 LinkedList234PalindromeLinkedList
Question: Why my code return false when comparing negative numbers like followings [-16557,-8725,-29125,28873,-21702,15483,-28441,-17845,-4317,-10914,-10914,-4317,-17845,-28441,15483,-21702,28873,-29125,-8725,-16557]
Answer: I use the ArrayList, Interger as on object, can only compare numbers between 2^7 to -2^7+1. So, for the object, we should use equals().
Further Question: ? How do we write equals() for Compare method?
203 RemoveLinkedListElements
1. Forgot move thecurr‘point’ to the next
when other if statement added
2. Usage of !=Null
We can use .next and .next.next to delete elements, but that means the current and next should be not null. ——> If we use .next.nextto for remove / delete, the curr = curr.nextshould be in the else statement
19 RemoveNthNodeFromEndOfList
自己第一反应做法是reverse list
1. Forgot the specialty of removing head element
2. if reverse you have to reverse back
看了一个solution的第一句话就晓得了通用办法，存在V2
1. 对于删除头结点的判断，嵌套的if else太过冗余，想想别的方法 4/15/2016
后来发现想多了。head = head.next 包含了head = null这种情况
//Create the two pointers ListNode faster = head; ListNode slower = head; for (int i = 0; i < n; i++) { faster = faster.next; }//Find the to-be-delete element (slower) if (faster == null) {//delete the head head = head.next; } else {//not delete head while (faster.next != null) { faster = faster.next; slower = slower.next; } slower.next = slower.next.next; }第一部分的for可以与第二部分while合并 - by网友先隔开n个再同时移动两个指针 ===> 一个循环中先数n个移动faster指针再一起移动两个指针
删除头结点的特殊性，其实可以在head前面再加个节点来解决！
LinkedList by Categories
总是忘了移动pointer
e.g.237 234 203 19PRint
忘记移动index这个point
关于reverse list究竟需要几个pointer
画一个图，模拟走两步，基本可以确定需要几个pointer。主要是头结点的特殊性，所以第一步的head算是特例。一般情况head不会作为pointer乱移动，所以宁可多设个pointer，不乱。忘记了删除头结点的特殊性
e.g. 19
比如LL19removeNthFromEnd, 在反向list里面有可能删除的是头结点.在没有用反向做的时候，当n == listlength 的时候也是删除头结点啊！判断 ！= null 的特殊性
同时判断.next != null和curr != null需要if else
