51 数组中重复的数字

哈希

时间On 空间On

bool duplicate(int numbers[], int length, int* duplication) {
    unordered_map<int, int> count;
    for (int i = 0; i < length; i++)
        count[numbers[i]]++;
    unordered_map<int, int>::iterator ite = count.begin();
    while (ite != count.end())
    {
        if (ite->second != 1)
        {
            *duplication = ite->first;
            return true;
        }
        ite++;
    }
    return false;
}

方法二

时间On空间O1

bool duplicate(int numbers[], int length, int* duplication)
{
    if (numbers == NULL || length <= 0)
        retrun false;
    for (int i = 0; i < length; i++)    //根据题意检查数字合法性
    {
        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]])    //找到重复数字
            {
                *duplication = numbers[i];
                return true;
            }
            swap(numbers[i], numbers[numbers[i]]); //放到该放的位置
        }
    }
    return false;
}