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.
快指针前移n-1步，如果fast->next==NULL；说明n=链表长度，head被删除；
否则，fast再前移一位，同时slow也前移，当fast到链尾时，slow->next就是倒数第n个节点；
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* removeNthFromEnd(struct ListNode* head, int n) { struct ListNode* fast = head; struct ListNode* slow = head; while(--n>0){ fast = fast->next; } if(fast->next == NULL){ return head->next; }else{ fast = fast->next; } while(fast->next != NULL){ fast = fast->next; slow = slow->next; } slow->next = slow->next->next; return head; }
leetcode 19. Remove Nth Node From End of List
原文：http://www.cnblogs.com/icelan/p/7760205.html