当if(num < k) 此时就表示查找失败 return 0;
while (p) {num == k} q = q->next; p = p->next;
如果num=k的话,此时说明p指针和q指针之间的差就是k-1,我们就让q和p指针一起往后移动,直到p遍历完链表。完整代码附上
//
// Created by 黎圣 on 2023/7/27.
//
#include "iostream"
typedef struct node
{
int data;
struct node *next;
}*linklist;
//我举的例子
int a[5] = {5,4,3,2,1};
int n = 5;
//创建链表L
void CreateList(linklist &L)
{
L = (linklist)malloc(sizeof(struct node));
struct node *s, *r = L;
for (int i = 0; i < n; i++)
{
s = (linklist)malloc(sizeof(struct node));
s->data = a[i];
r->next = s;
r = r->next;
}
s->next = NULL;
}
//打印出来方便观察
void display(linklist L)
{
struct node *s = L->next;
while (s)
{
printf("%d ", s->data);
s = s->next;
}
printf("\n");
}
//找倒数第k个元素
int findk(linklist L, int k)
{
struct node *p = L->next, *q = L->next;
int num = 0;
while (p)
{
if (num < k) num++;
else if (num == k) q = q->next;
p = p->next;
}
if (num < k) return 0;//不可能
else
{
printf("%d ", q->data);
return 1;
}
}
int main()
{
linklist L;
CreateList(L);
display(L);
findk(L, 2);//假设k值是2
}
这题的代码比较简单,重要的是找出差值是k-1这个规律,最后感谢b站up主@吸血小金鱼,思路都来自于她