[第二轮] 2/25-3/3 CareerCup 2.2

use two pointers front and back, the distance betwen these two pointers is K. so when the front pointer reaches the end, the back pointer is exactly at the location of kth to the last element.
Time complexity: O(n)
c/c++ 实现。初来咋到，望多多指教，谢谢大家，嘿嘿。
https://github.com/StanleyLee/CareerCup/blob/master/2_2.cpp

最开始连题都没理解。。。才知道kth to last是倒数第k个的意思。。。照着1楼的思路写的。。纯练手
java
https://gist.github.com/krisliu/5052641

Create two 'pointers', first one and second one have k nodes in between. Then go over the list, until the second one reaches the end. Same ideas as most of above.

https://gist.github.com/lynntian/5053658

jimwallet 发表于 2013-2-24 22:02
https://gist.github.com/njcongtou/4999347

1。 use two pointers, k interval between them.

you should juge n ,if n <= 0  

mapleman 发表于 2013-2-25 02:10
这道题二楼所说的那种做法，先遍历一遍拿到长度再去找倒数第k个，比答案差在哪里呢？
谢谢！

如果是n + n  的复杂度

Java language.
1.简单直接的办法，先遍历求长度，再求倒数k个元素
2.两个指针，距离为k，这还是参考了前面的同学，果然基础还是不行
3.其实我觉得将链表逆序然后反向求也是可以的，不过时间复杂度也没有优化，只能算一种思路吧。

https://gist.github.com/Sayary/5056257

find the kth to last element"equals" find N(Linked list lenth)-kth+1 to the first element. One pointer count the N, the other pointer point to N-K+1

https://gist.github.com/longwei/5059205

@Fanyare recursive的办法 优雅在于利用了function stack

https://gist.github.com/flatlight/5069975
用两个指针和遍历算长度再遍历不一样的吗...