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.
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* removeNthFromEnd(struct ListNode* head, int n) { if(head == NULL) return head; struct ListNode *pre = head; int i; for(i =0;i < n; i++) { if(pre == NULL) return NULL; pre = pre->next; } if(pre == NULL) return head->next; struct ListNode *last = head; while(pre->next) { pre = pre->next; last = last->next; } last->next = last->next->next; return head; }
leetcode - Remove Nth Node From End of List
原文：http://www.cnblogs.com/neyer/p/4612416.html