45 环形链表

0,1,2,...，n-1这n个数排列成一个圆圈，从数字0开始每次从这个圆圈里删除第m个数字。求这个圆圈里最后剩下的一个数字。

#include <iostream>
#include <list>

using namespace std;

int LastRemaining_Solution1(unsigned int n, unsigned int m)
{
    if(n < 1 || m <1)
    {
        return -1;
    }

    list<int> numbers;
    for (int i = 0; i < n; ++i)
    {
        numbers.push_back(i);
    }

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

        list<int>::iterator next = ++current;
        if(next == numbers.end())
            next = numbers.begin();

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

    return *(current);
}

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 != NULL)
        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(void)
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();
    Test6();

    return 0;
}

结果

QQ截图20160629214237.jpg