剑指offer---数组中重复的数字

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(length<=0||numbers==NULL)
            return false;
        //判断每一个元素是否非法
        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;
         
    }
};