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Question:
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.
Solution:
两个指针,距离相差n。下面的代码考虑了n异常的情况。代码在leetcode上做了验证。
/** * 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) { if (!head) return head; ListNode *node = new ListNode(0); node->next = head; ListNode *p1, *p2; p1 = p2 = node; for (int i = 0; i <n; i++) { p2 = p2->next; if (!p2) return head; } while (p2->next) { p2 = p2->next; p1 = p1->next; } p2 = p1->next; p1->next = p2->next; if (p2) delete p2; p1 = node->next; delete node; return p1; } };
Remove Nth Node From End of List
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原文地址:http://www.cnblogs.com/litao-tech/p/4330463.html