【第三轮】6.23-6.29 CareerCup 2.2

【解题思路】Use the runner technique. Use two pointers, move the runner k steps ahead
         * of the current pointer (if the runner reaches null it means k is larger
         * than the length, return null), then move them together one step at a
         * time. When the runner reaches the end the current points to the len-k th
         * element.
【时间复杂度】O(N)
【空间复杂度】O(1)
【gist link】https://gist.github.com/XiaoxiaoLi/d9b916553652fbeb6ef3

To Be Implemented

【解题思路】
  Two pointer point to head, first pointer move forward k elements, then move both pointer together, until the first one reaches the end, return second pointer.
【时间复杂度】
  O(N)
【空间复杂度】
  O(1)
【gist link】
  https://gist.github.com/habina/f521508cef301ac0d18d

【解题思路】Calculate the length of the list first, then get the (len-k-1)th element
【时间复杂度】O(N)
【空间复杂度】O(1)
【gist link】https://gist.github.com/nealhu/ff6cd5a1e217d47be280

jason51122 发表于 2014-6-25 13:53
To Be Implemented

【解题思路】Use 2 pointers. Keep the distance between slow and fast pointer to k. Move the windows to the end.
【时间复杂度】O(N)
【空间复杂度】O(1)
【gist link】https://gist.github.com/jason51122/2873201611ad2675b5a3

【解题思路】
2 pointers, the first one move k steps then move the pointers together till the the first one touch the end of the list
和 leetcode 上 Remove Nth Node From End of List 类似
【时间复杂度】
O(N)
【空间复杂度】
O(1)
【gist link】
https://gist.github.com/JoyceeLee/2fa8f900432f8759c79f

【解题思路】
  p1= head, p2指向第k个node。 一起遍历链表
【时间复杂度】
O(N)
【空间复杂度】
O(1)
【gist link】https://gist.github.com/yokiy/7da78a4396e307ced78d

【解题思路】2 pointers
【时间复杂度】O(N)
【空间复杂度】O(1)
【gist link】https://gist.github.com/pchong90/1f91c296ccfb81a8b639

【解题思路】 use two pointers p1 and p2, which are k nodes apart. Iterate the list and stop when p2 hits the end.  
【时间复杂度】O(N)
【空间复杂度】O(1)
【gist link】https://gist.github.com/jyhjuzi/1c751226bdbb31ea3e78