leetcode 19 Remove Nth Node From End of List
2015-05-05 22:31
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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.
解决方案:
Because the linked list have no knowledge about the previous nodes, we have to provide such information.
The difference between the final node and the to-be-delete node is N, hence we can utilize this information.
•front pointer points to the node which is N step away from the to-be-delete node
•rear pointer points to the to-be-delete node.
The algorithms is described as below:
•First driving front pointer N step forward.
•Secondly, move the 2 pointers 1 step ahead till the front pointer reach the end simultaneously, which will cause the rear pointer points to the previous node of the to-be-delete node.
•
Finally, jump the rear->next node by rear->next = rear->next->next.
下面的代码稍微有一个疑问:
http://bbs.csdn.net/topics/391029228
python解决方案:
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.
解决方案:
Because the linked list have no knowledge about the previous nodes, we have to provide such information.
The difference between the final node and the to-be-delete node is N, hence we can utilize this information.
•front pointer points to the node which is N step away from the to-be-delete node
•rear pointer points to the to-be-delete node.
The algorithms is described as below:
•First driving front pointer N step forward.
•Secondly, move the 2 pointers 1 step ahead till the front pointer reach the end simultaneously, which will cause the rear pointer points to the previous node of the to-be-delete node.
•
Finally, jump the rear->next node by rear->next = rear->next->next.
下面的代码稍微有一个疑问:
http://bbs.csdn.net/topics/391029228
class Solution { public: ListNode *removeNthFromEnd(ListNode *head, int n) { ListNode new_head(-1); new_head.next = head; ListNode *front = &new_head, *rear = &new_head; for (int i = 0; i < n; i++) front = front->next; while (front->next != NULL) { front = front->next; rear = rear->next; } ListNode *tmp = rear->next; rear->next = rear->next->next; delete tmp; head = new_head.next; return head; } };
python解决方案:
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: # @param {ListNode} head # @param {integer} n # @return {ListNode} def removeNthFromEnd(self, head, n): dummyHead = ListNode(0) dummyHead.next = head slow = fast = dummyHead for i in range(n): fast = fast.next while fast and fast.next: fast = fast.next slow = slow.next slow.next = slow.next.next return dummyHead.next
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