[leetcode] 19. 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.

解法一：

让cur先走n步，如果走到null，则说明需要移除head。 不然，再让cur和pre一起移动，当cur移动到null时，pre所在的node就是倒数第n个node。然后简单的操作pre->next即可。

/** * 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) {        if (!head) return head;        ListNode* pre= head, *cur = head;        for(int i = 0; i<n; ++i) cur = cur->next;        if(!cur) return head->next;                while(cur->next){            cur = cur->next;            pre = pre->next;        }        pre->next = pre->next->next;        return head;            }};