【第四轮】4.6 - 4.12 Career Cup 2.2

【solution】
use two pointers, firstly move fast runner k-1 steps forward, and then move two pointers both one step everytime until the fast runner hit the end, return the slow runner
【time】
O(n)
【space】
O(1)
【gist link 】
https://gist.github.com/kelly-us/69c2c202a0a85ae378b2

【解题思路】
# use two pointers, one move first for k - 1 step
# then move first and second pointer at the same time
# return the second value until the first node reaches the final spot
【时间复杂度】
O(n)
【空间复杂度】
O(1)
【gist link】
https://gist.github.com/zhangjiang2013/17431cb8a347cc104562

[solution]
use two pointers, make the fast pointer go k steps than the slow pointer
[time]
O(n)
[space]
O(1)
[gist]
https://gist.github.com/A30041839/0862e9279c7c14874e22

[solution]
1. slow and faster ptr. 2. recursion
[time]
1. O(n), 2. O(n)
[space]
1. O(1), 2.O(n)
[gist]
https://gist.github.com/songpu2015617/8ac85fd5ecb6b3cc3119/edit()

【解题思路】Iterative methods - use two pointers, one is kth ahead of the other one. Then forward these
        two pointers with the same pace. When the first pointer reach the end, the second pointer
        will point to the kth element to the last element
【时间复杂度】O(n)
【空间复杂度】O(1)
【gist link] https://github.com/StephenWeiXu/ ... aster/Chap2/2_2.cpp
【test case】(optional，如果觉得比较好，欢迎贴出来分享）

【解题思路】
use two pointers. Move runner pointer K steps. and then move walker and runner pointer simultaneously. When the runner hit the end of the linked list, return walker.
【时间复杂度】
O(n)
【空间复杂度】
O(1)

GistLink including test cases
https://gist.github.com/mogutou1214/8c3d5871ea1a7da4d234