剑指Offer-面试题3-数组中重复的数字

题目要求：

在一个长度为n的数组里的所有数字都在0到n-1的范围内。 数组中某些数字是重复的，但不知道有几个数字是重复的。也不知道每个数字重复几次。请找出数组中任意一个重复的数字。 例如，如果输入长度为7的数组{2,3,1,0,2,5,3}，那么对应的输出是第一个重复的数字2。

错误解法：

class Solution {
public:
    // Parameters:
    //        numbers:     an array of integers
    //        length:      the length of array numbers
    //        duplication: (Output) the duplicated number in the array number
    // Return value:       true if the input is valid, and there are some duplications in the array number
    //                     otherwise false
    bool duplicate(int numbers[], int length, int* duplication)
    {
        if （numbers == nullptr || length <= 0）
        {
            return false;
        }

        for (int i = 0; i < length; ++i)
        {
            if (numbers[i] < 0 || numbers[i] > length-1)
            {
                return false;
            }
        }

        for(int i=0; i<length; ++i)
        {
            while(numbers[i]!=i)
            {
                if(numbers[i]==numbers[numbers[i]])
                {
                    *duplication == numbers[i]；
                    return true;
                }

            //swap
                int temp = numbers[i];
                numbers[i] = numbers[temp];
                numbers[temp]= temp;
           }
        }
        return false;
    }
};

错误提示：

我还没有想明白

正确解法1：

class Solution {
public:
    // Parameters:
    //        numbers:     an array of integers
    //        length:      the length of array numbers
    //        duplication: (Output) the duplicated number in the array number
    // Return value:       true if the input is valid, and there are some duplications in the array number
    //                     otherwise false
    bool duplicate(int numbers[], int length, int* duplication)
{
    if(numbers == nullptr || length <= 0)
        return false;

    for(int i = 0; i < length; ++i)
    {
        if(numbers[i] < 0 || numbers[i] > length - 1)
            return false;
    }

    for(int i = 0; i < length; ++i)
    {
        while(numbers[i] != i)
        {
            if(numbers[i] == numbers[numbers[i]])
            {
                *duplication = numbers[i];
                return true;
            }

            // 交换numbers[i]和numbers[numbers[i]]             
            int temp = numbers[i];
            numbers[i] = numbers[temp];
            numbers[temp] = temp;
        }
    }

    return false;
}
};

该解法是书上给出的解法，采用“重排”思想实现，具体参考：

https://github.com/zhedahht/CodingInterviewChinese2/blob/master/03_01_DuplicationInArray/FindDuplication.cpp

正确解法2：

class Solution {

public:

    // Parameters:

    //        numbers:     an array of integers

    //        length:      the length of array numbers

    //        duplication: (Output) the duplicated number in the array number

    // Return value:       true if the input is valid, and there are some duplications in the array number

    //                     otherwise false

    bool duplicate(int numbers[], int length, int* duplication) {

        if(numbers==NULL||length==0) return 0;

        int hashTable[255]={0};

        for(int i=0;i<length;i++)

        {

            hashTable[numbers[i]]++;

        }

        int count=0;

        for(int i=0;i<length;i++)

        {

            if(hashTable[numbers[i]]>1)

            {

                duplication[count++]=numbers[i];

                //break;

                return true;

            }

        }

        return false;

    }

};

采用hashtable的思想实现，值得学习