LeetCode 19. Remove Nth Node From End of List
原题链接在这里：https://leetcode.com/problems/remove-nth-node-from-end-of-list/
题目：
Given a linked list, remove the nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
题解：
Method 1:算出总长度, 再减去n, 即为要从头多动的点。但要新要求, only one pass.
Method 2:两个指针，一个runner, 一个walker, runner先走n步，随后runner和walker一起走，直到runner指为空。
Note:1. 都是找到要去掉点的前一个点记为mark, 再通过mark.next = mark.next.next去掉对应应去掉的点.
2. 注意去掉list头点的情况，所以要建立一个dummy head. e.g. 1->2, n = 2.
Time Complexity: O(n), n是list长度. Space: O(1).
AC Java:
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution { 10 public ListNode removeNthFromEnd(ListNode head, int n) { 11 /*Method 1 12 if(head == null) 13 return head; 14 int len = 0; 15 ListNode temp = head; 16 while(temp != null){ 17 temp = temp.next; 18 len++; 19 } 20 if(len < n){ 21 return head; 22 } 23 int moveNum = len - n; 24 ListNode dunmy = new ListNode(0); 25 dunmy.next = head; 26 temp = dunmy; 27 while(moveNum > 0){ 28 temp = temp.next; 29 moveNum--; 30 } 31 temp.next = temp.next.next; 32 return dunmy.next; 33 */ 34 35 //Method 2 36 if(head == null || n == 0) 37 return head; 38 39 ListNode dummy = new ListNode(0); 40 dummy.next = head; 41 ListNode walker = dummy; 42 ListNode runner = dummy; 43 44 while(n>0 && runner.next != null){ 45 runner = runner.next; 46 n--; 47 } 48 49 while(runner.next != null){ 50 walker = walker.next; 51 runner = runner.next; 52 } 53 54 walker.next = walker.next.next; 55 return dummy.next; 56 } 57 }
AC C++:
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode() : val(0), next(nullptr) {} 7 * ListNode(int x) : val(x), next(nullptr) {} 8 * ListNode(int x, ListNode *next) : val(x), next(next) {} 9 * }; 10 */ 11 class Solution { 12 public: 13 ListNode* removeNthFromEnd(ListNode* head, int n) { 14 if(!head || n <= 0){ 15 return head; 16 } 17 18 ListNode* dummy = new ListNode(-1, head); 19 ListNode* walker = dummy; 20 ListNode* runner = dummy; 21 while(n > 0 && runner->next){ 22 runner = runner->next; 23 n--; 24 } 25 26 while(runner->next){ 27 walker = walker->next; 28 runner = runner->next; 29 } 30 31 walker->next = walker->next->next; 32 return dummy->next; 33 } 34 };
AC Python:
1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, val=0, next=None): 4 # self.val = val 5 # self.next = next 6 class Solution: 7 def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]: 8 if not head or n <= 0: 9 return head 10 11 dummy = ListNode(-1, head) 12 walker = runner = dummy 13 while n > 0 and runner.next: 14 runner = runner.next 15 n -= 1 16 17 while runner.next: 18 walker = walker.next 19 runner = runner.next 20 21 walker.next = walker.next.next 22 return dummy.next
类似Swapping Nodes in a Linked List, Delete the Middle Node of a Linked List.
