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.



/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        if (head == NULL) return NULL;
        if (n == 0) return head;
        ListNode *root = NULL;
        ListNode **index = &root;
        ListNode *j = head;


        while (head) {
            if (n != 0) {
                n--;
                head = head->next;
            } else { // n == 0
                *index = j;
                index = &(j->next);
                j = j->next;
                head = head->next;
            }
        }
        if (n == 0) {
            ListNode *tmp = j;
            *index = j->next;
            delete tmp;
        } else {
            *index = j;
        }
        return root;
    }

};

采用加dummy头的方法，注意不用new，可以直接用栈上的

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        ListNode prevHead(0);
        prevHead.next = head;
        
        ListNode *prev = &prevHead;
        ListNode *slow = head;
        ListNode *fast = head;
        
        for (int i = 0; i < n; i++)
            fast = fast->next;
            
        while (fast != NULL) {
            prev = prev->next;
            slow = slow->next;
            fast = fast->next;
        }
        
        prev->next = slow->next;
        delete slow;
        return prevHead.next;
    }
};