Remove Nth Node From End of List
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.
1 /************************************************************************* 2 > File Name: LeetCode019.c 3 > Author: Juntaran 4 > Mail: JuntaranMail@gmail.com 5 > Created Time: Tue 17 May 2016 15:59:22 PM CST 6 ************************************************************************/ 7 8 /************************************************************************* 9 10 Remove Nth Node From End of List 11 12 Given a linked list, 13 remove the nth node from the end of list and return its head. 14 15 For example, 16 17 Given linked list: 1->2->3->4->5, and n = 2. 18 19 After removing the second node from the end, 20 the linked list becomes 1->2->3->5. 21 22 Note: 23 Given n will always be valid. 24 Try to do this in one pass. 25 26 ************************************************************************/ 27 28 #include <stdio.h> 29 /** 30 * Definition for singly-linked list. 31 * struct ListNode { 32 * int val; 33 * struct ListNode *next; 34 * }; 35 */ 36 struct ListNode* removeNthFromEnd(struct ListNode* head, int n) 37 { 38 struct ListNode* fast = head; 39 struct ListNode* slow = head; 40 41 while( n > 0 ) 42 { 43 fast = fast->next; 44 n --; 45 } 46 if( fast == NULL ) 47 { 48 free(head); 49 return head->next; 50 } 51 while( fast->next != NULL ) 52 { 53 fast = fast->next; 54 slow = slow->next; 55 } 56 fast = slow->next; 57 slow->next = slow->next->next; 58 free(fast); 59 60 return head; 61 }