剑指offer之打印链表的倒数第N个节点的值

1 问题

打印链表的倒数第N个节点的值，（要求能只能便利链表一次）

比如链表如下，打印倒数第三个值就是4

1-> 2-> 3-> 4-> 5-> 6

2 思路

既然只要只能遍历一次，我们可以这样思考，比如我们要得到倒数第三个，那么它和尾巴的长度就是3，我们可以这一节距离一直往左边移动，那么移动最左边的话，他们的开始是1，尾巴是3，所以我们搞2个指针进行移动就行，如下过程，就可以得到4.

1 -> 2 -> 3 -> 4 -> 5 -> 6

start

end

1 -> 2 -> 3 -> 4 -> 5 -> 6

start end

1 -> 2 -> 3 -> 4 -> 5 -> 6

start end

无非就是上面的逆过程，我们用代码实现就行

3 代码实现

#include <iostream>using namespace std;typedef struct node
{int value;struct node *next;
} Node;void printN(Node *head, int n)
{if (head == NULL || n <= 0){std::cout << "head is NULL or n <= 0" << std::endl;return;	}Node *start = head;Node *end = head;//这里需要考虑n的大小长度是否大于链表长度//我们不能直接遍历链表得到链表大小然后和n比较//那我们就用start->next != NULL来判断也行for (int i = 0; i < n - 1; ++i){if (start->next != NULL)start = start->next;else{std::cout << "the value of n is more than larger the length of list" << std::endl;return;}}while (start->next != NULL){end = end->next;start = start->next;}std::cout << "the value is: " << end->value << std::endl;
}int main()
{Node *head = NULL;Node *node1 = NULL;Node *node2 = NULL;Node *node3 = NULL;head = (struct node*)malloc(sizeof(Node));node1 = (struct node*)malloc(sizeof(Node));node2 = (struct node*)malloc(sizeof(Node));node3 = (struct node*)malloc(sizeof(Node));	if (head == NULL || node1 == NULL || node2 == NULL || node3 == NULL){std::cout << "malloc fail" << std::endl;return -1;}head->value = 0;head->next = node1;node1->value = 1;node1->next = node2;node2->value = 2;node2->next = node3;node3->value = 3;node3->next = NULL;printN(head, 3);free(head);free(node1);free(node2);free(node3);return 0;}