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.
这题的难点在于one pass
没到尾部就没法进行倒退n个节点的操作。
但是到达尾部之后，再进行倒退删除操作，就不满足one pass了
由此诞生解法一：使用数组记录所有节点的位置，通过下标计算（尾节点位置-n+1）立刻定位到需要删除的节点了
建立vector存放ListNode*，每个指针指向一个链表节点
/** * 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) { vector<ListNode*> v; ListNode* cur = head; while(cur != NULL) { ListNode* p = cur; v.push_back(p); cur = cur->next; } int size = v.size(); int num = size-n; if(num == 0)//first, no pre return head->next; else if(num == size-1)//last but not first, no post { v[num-1]->next = NULL; return head; } else //pre link to post { v[num-1]->next = v[num+1]; return head; } } };
仔细分析之后觉得浪费空间太多。
我们只是通过尾节点位置确定需要删除节点（n-1个偏移量），不需要其他的位置信息。
只需要两个位置相差n-1的指针，当前面的指针指向尾节点时，后面的节点即指向需要删除的节点。
由此产生解法二：
/** * 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* fast = head; //the node at the last node ListNode* preslow = NULL; ListNode* slow = head; //the node to delete int i = 1; while(i < n) { fast = fast->next; i ++; } while(fast->next != NULL) { fast = fast->next; preslow = slow; slow = slow->next; } //fast at the last node, slow at the node to delete if(preslow == NULL) //delete the head return head->next; else { preslow->next = slow->next; return head; } } };
【LeetCode】Remove Nth Node From End of List (2 solutions)
原文：http://www.cnblogs.com/ganganloveu/p/4079879.html