Given a linked list, remove the n-th node from the end of list and return its head.
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.
Follow up:
Could you do this in one pass?
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这个题就是移去从表尾开始的第n 个元素。emmmm，这个最好画图，便于理解。
可以用双指针。
C++代码：
/** * 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 *s = head; ListNode *e = head; if(!head) return NULL; for(int i = 0; i < n; i++) e = e->next; if(!e) return head->next; while(e->next){ s = s->next; e = e->next; } s->next = s->next->next; return head; } };
转载于:https://www.cnblogs.com/Weixu-Liu/p/10704307.html
