19 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.

常规的解法，就是先轮训一次，计算其list的总长度len，再计算出需要要删除的index（len - n），但这需要两次循环。代码如下：

class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
int cnt = 0;
ListNode* node = head;
while (node) {
++cnt;
node = node->next;
}

int index = cnt - n;
if (0 == index) {
head = head->next;
} else {
node = head;
while (--index) {
node = node->next;
}
node->next = node->next->next;
}
return head;
}
};

上面的解法中，要先计算出len，才能算出index。我们可以用另外一种方法来计算出index，先循环n次，让node自增。再新建一个指针slow指向head，之后再接着node和slow自增，node为空时，slow也就刚好到了要删除的点了。代码如下：

ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode **slow = &head, *fast = head;
while (--n) fast = fast->next;
while(fast->next) {
slow = &((*slow)->next);
fast = fast->next;
}
*slow = (*slow)->next;
return head;
}