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.
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* removeNthFromEnd(struct ListNode* head, int n) { struct ListNode* p1 = head; struct ListNode* p2 = head; for(int i = 0; i < n; i++){ p1 = p1->next; } if(!p1){ //special condition: delete head head = head->next; free(p2); return head; } while(p1->next){ p1 = p1->next; p2 = p2->next; } p1 = p2->next; p2->next = p2->next->next; free(p1); return head; }
19. Remove Nth Node From End of List
原文：http://www.cnblogs.com/qionglouyuyu/p/5400712.html