JZ-C-45

剑指offer第四十五题：圆圈中最后剩下的数字：0,1,..,n-1这n个数排成一个圆圈，从数字0开始每次从圆圈中删除第m个数字。求出圆圈中剩下的最后一个数字（约瑟夫环★）

//============================================================================
// Name        : JZ-C-45.cpp
// Author      : Laughing_Lz
// Version     :
// Copyright   : All Right Reserved
// Description : 圆圈中最后剩下的数字：0,1,..,n-1这n个数排成一个圆圈，从数字0开始每次从圆圈中删除第m个数字。求出圆圈中剩下的最后一个数字（约瑟夫环★）
//============================================================================

#include <iostream>
#include <stdio.h>
#include <list>
using namespace std;

// ====================方法1====================
/**
 * 使用环形链表模拟圆圈，这里用模板库的std::list模拟环形链表，当迭代器扫描到链表末尾时候，将迭代器移到链表的头部
 * 每删除一个数字都需要m步运算，共n个数字，所以时间复杂度为O(m*n)，并且还需要一个辅助链表模拟圆圈，所以空间复杂度为O(n)
 */
int LastRemaining_Solution1(unsigned int n, unsigned int m) {
    if (n < 1 || m < 1)
        return -1;
    unsigned int i = 0;

    list<int> numbers;
    for (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) { //m也可以为1
            current++;
            if (current == numbers.end()) //当迭代器扫描到链表末尾时候，将迭代器移到链表的头部
                current = numbers.begin();
        }

        list<int>::iterator next = ++current;
        if (next == numbers.end()) //当迭代器扫描到链表末尾时候，将迭代器移到链表的头部
            next = numbers.begin();

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

    return *(current);
}

// ====================方法2====================
/**
 * n>1时：f(n,m) = [f(n-1,m)+m]%n;
 * n=1时：f(n,m) = 0;
 * 这种算法的时间复杂度为O(n)，空间复杂度为O(1)
 * 求解释！！！★★★
 */
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(char* testName, unsigned int n, unsigned int m, int expected) {
    if (testName != NULL)
        printf("%s begins: 
", testName);

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

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

    printf("
");
}

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();
    return 0;
}