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;
}