Remove Nth Node From End of List -- leetcode
2014-12-24 17:45
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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.
用一对指针;再引入一个假头,以统一处理节点删除操作。
以上算法是基于条件:Given n will always be valid.,即假定n输入都是符合预期的。
否则,还应该加上适当的边界处理。
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 fake(0); fake.next = head; ListNode *p = &fake; while (n--) p = p->next; ListNode *q = &fake; while (p->next) { p = p->next; q = q->next; } p = q->next; q->next = p->next; delete p; return fake.next; } };
用一对指针;再引入一个假头,以统一处理节点删除操作。
以上算法是基于条件:Given n will always be valid.,即假定n输入都是符合预期的。
否则,还应该加上适当的边界处理。
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