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.

解法一，雙指標大法：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(head == NULL || n <= 0)
            return head;

        ListNode* fast = head;
        for(int i=0; i<n && fast!=NULL; ++i)
            fast = fast->next;

        if(fast == NULL)
            return head->next;

        ListNode* slow = head;
        while(fast->next != NULL){
            slow = slow->next;
            fast = fast->next;
        }

        slow->next = slow->next->next;

        return head;
    }
};

解法二，stack+二級指標：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(head == NULL || n <= 0)
            return head;

        ListNode** pp = &head;
        std::stack<ListNode**> st;
        while(*pp != NULL){
            st.push(pp);
            pp = &((*pp)->next);
        }

        ListNode** target = NULL;
        int count = 0;
        while(!st.empty()){
            target = st.top();
            st.pop();
            if(++count == n)
                break;
        }

        *target = (*target)->next;

        return head;
    }
};

解法二可以通過測試用例，但是解法二沒有對N的長度是否大於連結串列長度進行判斷，不過如果N的長度=連結串列長度，解法二而*target=(*target)->next會滿足要求刪除了頭部。

解法一和二實際上都沒有處理連結串列長度K遠遠小於N的情況，實際上這種情況直接返回head就可以了。

所以解法一最完整的寫法應該是：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(head == NULL || n <= 0)
            return head;

        ListNode* fast = head;
        int i;
        for(i=0; i<n && fast!=NULL; ++i)
            fast = fast->next;

        if(fast == NULL){
            if(i == n)
                return head->next;
            else
                return head;
        }

        ListNode* slow = head;
        while(fast->next != NULL){
            slow = slow->next;
            fast = fast->next;
        }

        ListNode* tmp = slow->next;
        slow->next = slow->next->next;
        delete tmp;   //加上delete

        return head;
    }
};

解法二就不寫了。