// 面试题3（二）：不修改数组找出重复的数字// 题目：在一个长度为n+1的数组里的所有数字都在1到n的范围内，所以数组中至// 少有一个数字是重复的。请找出数组中任意一个重复的数字，但不能修改输入的// 数组。例如，如果输入长度为8的数组{2, 3, 5, 4, 3, 2, 6, 7}，那么对应的// 输出是重复的数字2或者3。#include 
     
      using namespace std;int counter(const int*, int, int, int);bool getDuplication(const int* numbers, int length, int& num){    if (numbers == NULL || length <= 0)//判断输入是否对        return false;    for (int i = 0; i < length; i++)//判断是否满足题目条件    {        if (numbers[i] < 1 || numbers[i] > length)            return false;    }    int start = 1, end = length - 1;    while (end >= start)    {        int middle = ((end - start) >> 1) + start;        int count = counter(numbers, length, start, middle);//查找落在二分左区间内个数        //cout << "start=" << start << endl << "middle=" << middle << endl << "end=" << end << endl<< "count=" << count << endl;                if (start == end)//二分不动了，停止，判断这个值count值        {            if (count > 1)            {                num = start;                return true;            }            else                break;                    }        if (count > (middle - start) + 1)//如果落在左区间的个数大于区间范围，则这里面一定有重复，否则就去右区间看看            end = middle;        else            start = middle + 1;    }    return false;}int counter(const int* numbers, int length, int start, int middle){    int count = 0;    if (numbers == NULL || start > middle || start < 0)        return count;    for (int i = 0; i < length; i++)    {        if (numbers[i] >= start&&numbers[i] <= middle)            count++;    }    return count;}void test(const char* testName, int* numbers, int length, int* duplications, int dupLength){    int result;    bool flag = getDuplication(numbers, length, result);    if (!flag)        result = -1;    for (int i = 0; i < dupLength; ++i)    {        if (result == duplications[i])        {            std::cout << testName << " passed." << std::endl;            return;        }    }    std::cout << testName << " FAILED." << std::endl;}// 多个重复的数字void test1(){    int numbers[] = { 2, 3, 5, 4, 3, 2, 6, 7 };    int duplications[] = { 2, 3 };    test("test1", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));}// 一个重复的数字void test2(){    int numbers[] = { 3, 2, 1, 4, 4, 5, 6, 7 };    int duplications[] = { 4 };    test("test2", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));}// 重复的数字是数组中最小的数字void test3(){    int numbers[] = { 1, 2, 3, 4, 5, 6, 7, 1, 8 };    int duplications[] = { 1 };    test("test3", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));}// 重复的数字是数组中最大的数字void test4(){    int numbers[] = { 1, 7, 3, 4, 5, 6, 8, 2, 8 };    int duplications[] = { 8 };    test("test4", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));}// 数组中只有两个数字void test5(){    int numbers[] = { 1, 1 };    int duplications[] = { 1 };    test("test5", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));}// 重复的数字位于数组当中void test6(){    int numbers[] = { 3, 2, 1, 3, 4, 5, 6, 7 };    int duplications[] = { 3 };    test("test6", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));}// 多个重复的数字void test7(){    int numbers[] = { 1, 2, 2, 6, 4, 5, 6 };    int duplications[] = { 2, 6 };    test("test7", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));}// 一个数字重复三次void test8(){    int numbers[] = { 1, 2, 2, 6, 4, 5, 2 };    int duplications[] = { 2 };    test("test8", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));}// 没有重复的数字void test9(){    int numbers[] = { 1, 2, 6, 4, 5, 3 };    int duplications[] = { -1 };    test("test9", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));}// 无效的输入void test10(){    int* numbers = nullptr;    int duplications[] = { -1 };    test("test10", numbers, 0, duplications, sizeof(duplications) / sizeof(int));}void main(){    test1();    test2();    test3();//判断边界    test4();//判断边界    test5();    test6();    test7();    test8();    test9();    test10();    system("pause");}