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.

Solution:

/**
 * 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) {
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        ListNode *p1 = dummy, *p2 = dummy;
        for ( int i = 0; i <= n; i++ ) p1 = p1->next;
        while ( p1 != NULL ) {
            p1 = p1->next;
            p2 = p2->next;
        }
        p2->next = p2->next->next;
        return dummy->next;
    }
};

Use the following examples to determine some of the boundary condition and corner case.

        /**
         * n = 2
         * dummy0     1  2  3  4  5  NULL
         *                  p1       p1'
         *      p2          p2'
         */
         /**
         * n = 5
         * dummy0     1  2  3  4  5  NULL
         *                           p1(')
         *      p2(')
         */

find n-th node in a linked list.

// find the n-th node in the linked list
ListNode *p1 = head;
for ( int i = 1; i <= n-1; i++ ) p1 = p1->next;

count the number of nodes in a linked list

// icount is the number of nodes in a linked list
ListNode *p3 = head;
int icount = 0;
while ( p3 != NULL ) {p3 = p3->next; icount += 1;}