微软等数据结构与算法面试100题 第十三题

第十三题

题目：输入一个单向链表，输出该链表中倒数第k个结点。

这道题比较简单，就是对于这个链表，定义两个指针head1 head2，然后让head1向前走k-1个位置以后，head2和head1同时向前走，知道head1知道NULL指针，head2的即为倒数第k个指针。

代码：

#include<iostream>
using namespace std;

typedef struct node
{
node * nodeNext;
int value;
}node;

int seekReListK(node *head, int k)
{
if(NULL==head)
{
cout<<"空链表";
return -1;
}
node *head1 = head;
node *head2 = head;
k = k - 1;
while(k)
{
k = k - 1;
if(NULL!=head1 ->nodeNext)
head1 = head1 ->nodeNext;
else
{
cout<<"K超过链表的长度"<<endl;
return -1;
}
}
//现在head1比head2快k个
while(NULL!=head1->nodeNext)
{
head1 = head1->nodeNext;
head2 = head2->nodeNext;
}

return head2->value;
}
int main()
{
//构建链表
node *a = new struct node();
node *b = new struct node();
node *c = new struct node();
node *d = new struct node();
node *e = new struct node();
node *f = new struct node();
node *g = new struct node();
node *h = new struct node();
node *i = new struct node();
node *j = new struct node();
a->nodeNext = b;
b->nodeNext = c;
c->nodeNext = d;
d->nodeNext = e;
e->nodeNext = f;
f->nodeNext = g;
g->nodeNext = h;
h->nodeNext = i;
i->nodeNext = j;
j->nodeNext = NULL;
a->value = 0;
b->value = 1;
c->value = 2;
d->value = 3;
e->value = 4;
f->value = 5;
g->value = 6;
h->value = 7;
i->value = 8;
j->value = 9;

cout<<seekReListK(a,1)<<endl;
cout<<seekReListK(a,10)<<endl;
cout<<seekReListK(a,11)<<endl;
return 0;
}