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.
题目要求我们要尽量扫描一遍，那我们就要设置两个指针了，第二个指针先前行n+1，然后第一个指针再前行，当第二个指针到达结尾时，第一个指针就是
要删除节点的前序，同时也要考虑恰好头结点就是要删除节点的特殊情况
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* removeNthFromEnd(ListNode* head, int n) { ListNode *pre = head, *current = head; int count = 0; while (current) { if (!(count <= n)) pre = pre->next; ++count; current = current->next; } if (pre == head&&count==n) { head = head->next; free(pre); } else { current = pre->next; pre->next = pre->next->next; free(current); } return head; } };
LeetCode 19. Remove Nth Node From End of List
原文：http://www.cnblogs.com/csudanli/p/5883458.html