Given a linked list, remove thenthnode 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:Givennwill always be valid.
Try to do this in one pass.
算法：
1 准备两个指针first, second
2 先让fisrt走n步
3 让fisrt和second同时走直到first遇到结尾
4 要用一个temp指针来记录second前一个node，用来删除second用。
/** * 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 *first = head, *second = head;ListNode *temp = second;for(int i = 0; i < n; i++) {first = first -> next;}while(first) {first = first -> next;temp = second;second = second -> next;}if(second == head) {head = head -> next;}else {temp -> next = second -> next;}return head;}};
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