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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.
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 || n < 1 ) return head; //若head为空，或n不合法13 ListNode node(0); //设置头结点，好删除14 node.next = head;15 ListNode* pre = &node; //slow的前驱节点16 ListNode* fast = head;17 while( n && fast ) { //令fast先指向第n个节点18 fast = fast->next;19 --n;20 }21 if( n && !fast ) return head; //若链表长度没有n大22 ListNode* slow = head;23 while( fast ) { //fast和slow同时后移，slow为空时结束24 fast = fast->next;25 pre = slow;26 slow = slow->next;27 }28 pre->next = slow->next; //删除第n个节点29 delete slow;30 return node.next;31 }32 };
设立前后两指针，假设为前指针为fast，后指针为slow，先让fast指针指向head，向后走n步，指向第n个节点，然后slow指针指向head，让fast指针和slow指针同时后移，知道slow为空，那么fast指针指向的就是倒数第n个节点
Remove Nth Node From End of List
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