給出連結串列 3->2->1->5->null和n = 2，返回倒數第二個節點的值1.

題目連結

程式

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */


class Solution {
public:
    /*
     * @param head: The first node of linked list.
     * @param n: An integer
     * @return: Nth to last node of a singly linked list. 
     */
    ListNode * nthToLast(ListNode * head, int n) {
        // write your code here
        if(head == NULL)
            return head;
        ListNode *cur = head;
        int size = 0;
        //找到連結串列的長度
        while(cur){
            cur = cur->next;
            size++;
        }
        //從前向後數 
        cur = head;
        for(int i = 0; i < size - n; i++){
            cur = cur->next;
        }
        return cur;
    }
};