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[LeetCode] 19. Remove Nth Node From End of List
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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