《剑指offer》第六十二题（圆圈中最后剩下的数字）

// 面试题62：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。

#include <iostream>
#include <list>

using namespace std;

// ====================方法1====================
//使用环形链表
int LastRemaining_Solution1(unsigned int n, unsigned int m)
{
    if (n < 1 || m < 1)//边界判断
        return -1;

    unsigned int i = 0;

    list<int> numbers;//建立一个链表，值为0~n
    for (i = 0; i < n; ++i)
        numbers.push_back(i);

    list<int>::iterator current = numbers.begin();//设置迭代器
    while (numbers.size() > 1)
    {
        for (int i = 1; i < m; ++i)//找到待删除节点
        {
            current++;
            if (current == numbers.end())//遇到尾节点，设置成头节点
                current = numbers.begin();
        }

        list<int>::iterator next = ++current;//将当前待删除的节点的下个节点作为开始
        if (next == numbers.end())
            next = numbers.begin();

        --current;//因为上面有++，所以这里--
        numbers.erase(current);//删除当前节点
        current = next;
    }

    return *(current);//上面循环完就剩一个了
}

// ====================方法2====================
//需要严谨的数学公式推导，最终得到如下的公式
int LastRemaining_Solution2(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;
}

// ====================测试代码====================
void Test(const char* testName, unsigned int n, unsigned int m, int expected)
{
    if (testName != nullptr)
        printf("%s begins: \n", testName);

    if (LastRemaining_Solution1(n, m) == expected)
        printf("Solution1 passed.\n");
    else
        printf("Solution1 failed.\n");

    if (LastRemaining_Solution2(n, m) == expected)
        printf("Solution2 passed.\n");
    else
        printf("Solution2 failed.\n");

    printf("\n");
}

void Test1()
{
    Test("Test1", 5, 3, 3);
}

void Test2()
{
    Test("Test2", 5, 2, 2);
}

void Test3()
{
    Test("Test3", 6, 7, 4);
}

void Test4()
{
    Test("Test4", 6, 6, 3);
}

void Test5()
{
    Test("Test5", 0, 0, -1);
}

void Test6()
{
    Test("Test6", 4000, 997, 1027);
}

int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();
    Test6();
    system("pause");
    return 0;
}