LeetCode: Remove Nth Node From End of List

Title：

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.


思路： 先让一个指针走n步，在让另一个指针一起向前。需要注意n的取值。有可能直接删除头指针，需要判断

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        if(head == NULL){
            return head;
        }
        ListNode* dummy = new ListNode(0); // 处理删除头指针
        dummy->next = head;
        head = dummy;
        ListNode* slow = head;
        ListNode* fast = head;
        for(int i = 0; i <n; i++){
            fast = fast ->next;
        }
        while(fast->next != NULL){
            fast = fast->next;
            slow = slow->next;
        }
        ListNode* temp = slow->next;
        slow ->next = slow->next->next;
        delete temp;
        return dummy->next;
    }
};