(剑指Offer)面试题45:圆圈中最后剩下的数字

题目:

0,1,…n-1这n个数字排成一个圆圈,从数字0开始每次从这个圆圈里删除第m个数字,求出这个圆圈里剩下的最后一个数字。

思路:

1、环形链表模拟圆圈

创建一个n个节点的环形链表,然后每次在这个链表中删除第m个节点;

可以用std::list来模拟环形链表,list本身不是环形结构,因此每当迭代器扫描到链表末尾的时候,需要将迭代器移到链表的头部。

2、分析每次被删除的数字的规律,动态规划

假设从0-n-1中删除了第m个数字,则下一轮的数字排列为m,m+1,…..n,1,2,3…m-2,将该数字排列重新映射为0~n-2,则为

m    0

m+1    1  

….    ….

n-1   n-1-m

0    n-m

1    n-m+1

…    ….

m-2    n-2

可以看出从右往左的映射关系为left=(right+m)%n,即0~n-1序列中最后剩下的数字等于(0~n-2序列中最后剩下的数字+m)%n,很明显当n=1时,只有一个数,那么剩下的数字就是0.

问题转化为动态规划问题,关系表示为:

f(n)=(f(n-1)+m)%n; 当n=1,f(1)=0;

代码:

1、环形链表

#include <iostream>
#include <list>
using namespace std;

int lastRemaining(unsigned int n,unsigned int m){
    if(n<1 || m<1)
        return  -1;

    std::list<int> numbers;
    for(unsigned int i=0;i<n;i++)
        numbers.push_back(i);

    std::list<int>::iterator current=numbers.begin();
    std::list<int>::iterator next;
    while(numbers.size()>1){
        for(unsigned int i=1;i<m;i++){
            current++;
            if(current==numbers.end())
                current=numbers.begin();
        }
        next=++current;
        if(next==numbers.end())
            next=numbers.begin();

        --current;
        numbers.erase(current);
        current=next;
    }
    return *(current);
}

int main()
{
    cout << lastRemaining(5,3) << endl;
    return 0;
}

2、映射规律,动态规划

int lastRemaining_1(unsigned int n,unsigned int m){
    if(n<1 || m<1)
        return  -1;

    int last=0;
    for(int i=2;i<=n;i++){
        last=(last+m)%i;
    }
    return last;
}


int lastRemaining_2(unsigned int n,unsigned int m){
    if(n<1 || m<1)
        return  -1;
    if(n==1)
        return 0;
    return (lastRemaining_2(n-1,m)+m)%n;
}

在线测试OJ:

http://www.nowcoder.com/books/coding-interviews/f78a359491e64a50bce2d89cff857eb6?rp=2

AC代码:

1、环形链表

class Solution {
public:
    int LastRemaining_Solution(unsigned int n, unsigned int m)
    {
    	if(n<1 || m<1)
            return -1;

        std::list<int> numbers;
        for(unsigned int i=0;i<n;i++)
            numbers.push_back(i);

        std::list<int>::iterator current=numbers.begin();
        std::list<int>::iterator next;

        while(numbers.size()>1){
            for(unsigned int i=1;i<m;i++){
                ++current;
                if(current==numbers.end())
                    current=numbers.begin();
            }

            next=++current;
            if(next==numbers.end())
                next=numbers.begin();

            --current;
            numbers.erase(current);
            current=next;
        }

        return *current;
    }
};

2、序列规律,动态规划

class Solution {
public:
    int LastRemaining_Solution(unsigned int n, unsigned int m)
    {
        if(n<1 || m<1)
            return -1;
        int last=0;
        for(int i=2;i<=n;i++)
            last=(last+m)%i;
        
        return last;
    }
};

class Solution {
public:
    int LastRemaining_Solution(unsigned int n, unsigned int m)
    {
        if(n<1 || m<1)
            return -1;
        if(n==1)
            return 0;
        
        return (LastRemaining_Solution(n-1,m)+m)%n;
    }
};
    原文作者:AndyJee
    原文地址: https://www.cnblogs.com/AndyJee/p/4687715.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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