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题目要求：
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.
这道题比较常规的想法是先算出链表长度len，再将头指针移动len-n步就可以到达待删节点了。具体程序如下（4ms）：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* removeNthFromEnd(ListNode* head, int n) { 12 if(!head) 13 return head; 14 15 int len = 0; 16 ListNode *start = head; 17 while(start) 18 { 19 len++; 20 start = start->next; 21 } 22 23 if(len == n) 24 { 25 start = head; 26 head = head->next; 27 delete start; 28 start = nullptr; 29 return head; 30 } 31 32 ListNode *preNode = head; 33 start = preNode->next; 34 int k = 1; 35 while(k < len - n) 36 { 37 preNode = start; 38 start = start->next; 39 k++; 40 } 41 42 preNode->next = start->next; 43 delete start; 44 start = nullptr; 45 46 return head; 47 } 48 };
另一个比较巧妙的方法就是先定义一个慢指针（初始值为头指针父指针），让头指针移动n步，再让头指针和一个慢指针同时移动直至头指针为空，这时慢指针指向的下一个节点就是待删节点。具体程序日如下（4ms）：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* removeNthFromEnd(ListNode* head, int n) { 12 if(!head) 13 return head; 14 15 ListNode *dummy = new(nothrow) ListNode(INT_MIN); 16 assert(dummy); 17 dummy->next = head; 18 19 for(int i = 0; i < n; i++) 20 head = head->next; 21 22 ListNode *preNode = dummy; 23 while(head) 24 { 25 head = head->next; 26 preNode = preNode->next; 27 } 28 29 ListNode *delNode = preNode->next; 30 preNode->next = delNode->next; 31 delete delNode; 32 delNode = nullptr; 33 34 head = dummy->next; 35 delete dummy; 36 dummy = nullptr; 37 38 return head; 39 } 40 };
LeetCode之“链表”：Remove Nth Node From End of List
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原文地址：http://www.cnblogs.com/xiehongfeng100/p/4603750.html
