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leetcode - Remove Nth Node From End of List
题目：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.
个人思路：
1、设置两个指针，一个快，一个慢，根据给定的n，让快指针先走n步，之后再同时走，直到快指针走到末节点，此时慢指针走到要删除节点的父节点
2、注意一下边界条件即可，链表为空、删除头节点等情况
代码：
1 #include <stddef.h> 2 3 struct ListNode 4 { 5 int val; 6 ListNode *next; 7 ListNode(int x) : val(x), next(NULL) {}; 8 }; 9 10 class Solution11 {12 public:13 ListNode* removeNthFromEnd(ListNode *head, int n)14 {15 if (!head)16 {17 return NULL;18 }19 20 ListNode *slow = head;21 ListNode *fast = head;22 23 //快指针先走n步24 for (int i = 0; i < n; ++i)25 {26 fast = fast->next;27 }28 //快指针为空，表明要删除的节点为头节点29 if (!fast)30 {31 head = head->next;32 delete slow;33 slow = NULL;34 35 return head;36 }37 //快指针走到最后一个节点时，慢指针走到要删除节点的前一个节点38 while (fast->next)39 {40 slow = slow->next;41 fast = fast->next;42 }43 //删除节点44 ListNode *deleted = slow->next;45 slow->next = deleted->next;46 delete deleted;47 deleted = NULL;48 49 return head;50 }51 };52 53 int main()54 {55 return 0;56 }
网上的文章大部分都是这个思路
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