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.

思路：遍历链表，使用数组保存每个节点的地址。

class Solution {
public:
ListNode *removeNthFromEnd( ListNode *head, int n ) {
if( !head ) { return 0; }
vector<ListNode*> nodeVec;
int size = 0;
while( head ) {
nodeVec.push_back( head );
head = head->next;
++size;
}
if( size-n == 0 ) { return nodeVec[0]->next; }
nodeVec[size-n-1]->next = nodeVec[size-n]->next;
return nodeVec[0];
}
};

网上看到的思路，使用两个指针遍历链表，两个指针相距n。对于倒数第n个元素，与其相距n的元素将为空。因此，当第二指针为空时，第一个指针指向的即为要删除的元素。

class Solution {
public:
ListNode *removeNthFromEnd( ListNode *head, int n ) {
ListNode guard(-1);
guard.next = head;
head = &guard;
for( int i = 0; i < n; ++i ) { head = head->next; }
ListNode *node = &guard;
while( head->next ) {
node = node->next;
head = head->next;
}
node->next = node->next->next;
return guard.next;
}
};