《剑指offer》第三_二题

Posted 2021-02-11 cjt-blog

// 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 <iostream>
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");
}