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，双指针大法：

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+二级指针：

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;    }};

解法二可以通过测试用例，但是解法二没有对N的长度是否大于链表长度进行判断，不过如果N的长度=链表长度，解法二而*target=(*target)->next会满足要求删除了头部。

解法一和二实际上都没有处理链表长度K远远小于N的情况，实际上这种情况直接返回head就可以了。

所以解法一最完整的写法应该是：

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;    }};

解法二就不写了。