[LeetCode] Remove Nth Node From End of List
2014-05-09 19:32
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Given a linked list, remove the nth node from the end of list and return its head.
For example,
Note:
Given n will always be valid.
Try to do this in one pass.
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; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *removeNthFromEnd(ListNode *head, int n) { ListNode *p = head, *q = head, *pre = nullptr; for(int i = 0; i < n-1; i++) q = q->next; while(q->next) { pre = p; p = p->next; q = q->next; } if(pre == nullptr) { head = p->next; delete p; } else { pre->next = p->next; delete p; } return head; } };
/** * 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 dummy(0); dummy.next = head; ListNode *p = &dummy, *q = &dummy; while(n--) q = q->next; while(q->next) { p = p->next; q = q->next; } ListNode *tmp = p->next; p->next = tmp->next; delete tmp; return dummy.next; } };
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