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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.
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* removeNthFromEnd(ListNode* head, int n) { 12 ListNode *dummy = new ListNode(0); 13 dummy->next = head; 14 ListNode *ret = NULL; 15 ListNode *prev = dummy; 16 17 ListNode *p1 = head; 18 ListNode *p2 = head; 19 int counter = 0; 20 21 while (p2){ 22 p2 = p2->next; 23 counter++; 24 if (counter == n){ 25 break; 26 } 27 } 28 29 if (counter < n){ 30 delete dummy; 31 return head; 32 } 33 34 while(p2){ 35 p1 = p1->next; 36 p2 = p2->next; 37 prev = prev->next; 38 } 39 40 prev->next = p1->next; 41 42 ret = dummy->next; 43 delete dummy; 44 return ret; 45 } 46 };
[LeetCode] 19. Remove Nth Node From End of List
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原文地址:http://www.cnblogs.com/amadis/p/5926516.html