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.
各种情况没想全，重复提交了很多次，需要再更加仔细的思考……
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/** * 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
i = 0; ListNode *p = head; for(i = 0; i < n-1 ; i++){ //notice second node from the end just one step from the end p = p->next; } ListNode *r = NULL; //r is the poiter ahead of q ListNode *q = head; while(p->next != NULL){ p = p->next; r = q; q = q->next; } if(r == NULL) head = head->next; else{ r->next = q->next; } return
head; } }; |
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Remove Nth Node From End of List
原文：http://www.cnblogs.com/pengyu2003/p/3572437.html