lintcode-166-链表倒数第n个节点

166-链表倒数第n个节点

找到单链表倒数第n个节点,保证链表中节点的最少数量为n。

样例

给出链表 3->2->1->5->null和n = 2,返回倒数第二个节点的值1.

标签

链表 Cracking The Coding Interview

思路

使用 快慢个指针,快指针比慢指针先走 n 步,当快指针走到链表尾部时,慢指针指向倒数第 n 个节点

code

/**
 * 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
        ListNode *p1 = head, *p2 = head;
        for (int i = 0; i < n; i++) {
            p1 = p1->next;
        }
        while (p1 != NULL) {
            p1 = p1->next;
            p2 = p2->next;
        }
        return p2;
    }
};
原文地址:https://www.cnblogs.com/libaoquan/p/7275674.html