剑指offer-圆周中最后剩下的数字

/*******************************************************************
Copyright(c) 2016, Tyrone Li
All rights reserved.
*******************************************************************/
// 作者:TyroneLi
//
 
/*
Q1:
    圆圈中最后剩下的数字:
        0,1,2,...,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2中删
        除第m个数字，求出这个圆圈里剩下的最后一个数字。（约瑟夫环问题）
S1:
    1. 用环形链表模拟圆圈解决，思路直接，但是需要耗费一定的时间效率和空间效率。
       时间复杂度O(nm)，空间复杂度O(n);
    2. 分析每次删除的数字以及剩下的序列的规律，直接找到最后所剩下的数字。
        时间复杂度O(n), 空间复杂度O(1);更好
*/

#include 
#include 
#include 
#include 
#include 
#include 
#include 

int lastRemaining(unsigned int n, unsigned int m)
{
    if(n < 1 || m < 1)
        return -1;
    
    unsigned int i = 0;
    std::list data;
    for(i = 0; i < n; ++i)
        data.push_back(i);
    
    std::list::iterator curIter = data.begin();
    // 当链表的指针指向最后一个元素的下一个元素时候，讲迭代器
    // 重新指向第一个元素，从而保证可以模拟环状的链表。
    while(data.size() > 1)
    {
        for(int i = 1; i < m; ++i)
        {
            curIter++;
            if(curIter == data.end())
                curIter = data.begin();
        }
        std::list::iterator nextIter = (++curIter);
        if(nextIter == data.end())
            nextIter = data.begin();
        --curIter;
        data.erase(curIter);
        curIter = nextIter;
    }
    return (*curIter);
}

int lastRemaining_quick(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;
}

void test_1()
{
    std::cout << "Test 1" << std::endl;
    std::cout << "Expect : " << 3 << " Get : " << lastRemaining(5, 3) << std::endl;
    std::cout << "Expect : " << 3 << " Get : " << lastRemaining_quick(5, 3) << std::endl;
    std::cout << std::endl;
}

void test_2()
{
    std::cout << "Test 2" << std::endl;
    std::cout << "Expect : " << 0 << " Get : " << lastRemaining(1, 3) << std::endl;
    std::cout << "Expect : " << 0 << " Get : " << lastRemaining_quick(1, 3) << std::endl;
    std::cout << std::endl;
}

void test_3()
{
    std::cout << "Test 3" << std::endl;
    std::cout << "Expect : " << 2 << " Get : " << lastRemaining(5, 2) << std::endl;
    std::cout << "Expect : " << 2 << " Get : " << lastRemaining_quick(5, 2) << std::endl;
    std::cout << std::endl;
}

void test_4()
{
    std::cout << "Test 4" << std::endl;
    std::cout << "Expect : " << 3 << " Get : " << lastRemaining(6, 3) << std::endl;
    std::cout << "Expect : " << 3 << " Get : " << lastRemaining_quick(6, 3) << std::endl;
    std::cout << std::endl;
}

void test_5()
{
    std::cout << "Test 5" << std::endl;
    std::cout << "Expect : " << 4 << " Get : " << lastRemaining(6, 4) << std::endl;
    std::cout << "Expect : " << 4 << " Get : " << lastRemaining_quick(6, 4) << std::endl;
    std::cout << std::endl;
}

void test_6()
{
    std::cout << "Test 6" << std::endl;
    std::cout << "Expect : " << -1 << " Get : " << lastRemaining(0, 0) << std::endl;
    std::cout << "Expect : " << -1 << " Get : " << lastRemaining_quick(0,0) << std::endl;
    std::cout << std::endl;
}

void test_7()
{
    std::cout << "Test 7" << std::endl;
    std::cout << "Expect : " << 1027 << " Get : " << lastRemaining(4000, 997) << std::endl;
    std::cout << "Expect : " << 1027 << " Get : " << lastRemaining_quick(4000, 997) << std::endl;
    std::cout << std::endl;
}

void test_lastRemaining()
{
    test_1();
    test_2();
    test_3();
    test_4();
    test_5();
    test_6();
    test_7();
}

int main(int argc, char**argv)
{

    test_lastRemaining();

    return 0;
}