題目描述

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)
    {
        map<int
 
, int> mapping;

        for (int i = 0; i < length; i++)
        {
            if (mapping[numbers[i]])
            {
                *duplication = numbers[i];
                return true;
            }
            else
                mapping[numbers[i]] += 1;
        }
        return false;
    }
};
 

解法二：

因為題目指明陣列內陣列在0~n-1之間，則對每個元素，都增加其元素作為索引所在位置上的值到超過length。當碰到下一次同樣元素時，判斷該值大於length，說明出現過一次，則返回該元素。

時間o(n) ，空間o(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)
    {
        for (int i = 0; i < length; i++)
        {
            int index = numbers[i] % length;
            if (numbers[index] < length)
            {
                numbers[index] += length;
            }
            else
            {
                *duplication = numbers[index] - length;
                return true;
            }
        }
        return false;
    }
};