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leetcode第19题--Remove Nth Node From End of List
Problem：
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.
果然一次就通过了。先读到尾，知道长度后，再从头开始读到要去掉的前一位，然后把要去掉的next复制给要去掉的前一个的next就可以。
/** * 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){ int len = 1; ListNode *tmp = head; while(tmp -> next != NULL) { len++; tmp = tmp -> next; } if (len == n) return head -> next; ListNode *p = head; for(int i = 1; i < len - n; i++) { p = p -> next; } p -> next = p -> next -> next; return head;}};
leetcode第19题--Remove Nth Node From End of List
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