H-Index && H-Index II

Given an array of citations (each citation is a non-negative integer) of a researcher, write a function to compute the researcher‘s h-index.

According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers have at least h citations each, and the other N ? h papers have no more than h citations each."

For example, given citations = [3, 0, 6, 1, 5], which means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 citations respectively. Since the researcher has 3 papers with at least 3 citations each and the remaining two with no more than 3 citations each, his h-index is 3.

Note: If there are several possible values for h, the maximum one is taken as the h-index.

至少有h篇被人引用h次

• 如果总共的论文数为N则，h最多为N
• 计数排序

int hIndex(int* citations, int citationsSize) {
    int *array = (int *)malloc((citationsSize + 1) * sizeof(int));
    for (int i = 0; i < citationsSize + 1; i++)
        array[i] = 0;        //必须初始化
    for (int i = 0; i < citationsSize; i++)
    {
        if (citations[i] >= citationsSize)    //如果大于最大文章数，则认为最大引用文章值++
            array[citationsSize]++;
        else
            array[citations[i]]++;
    }
    if (array[citationsSize] >= citationsSize)
        return citationsSize;
    for (int i = citationsSize - 1; i >= 0; i--)
    {
        array[i] = array[i] + array[i + 1];
        if (array[i] >= i)
            return i;
    }
    return 0;
}

Follow up for H-Index: What if the citations array is sorted in ascending order? Could you optimize your algorithm?

int hIndex(int* citations, int citationsSize) {
    int n = 0;
    if(citations[0] >= citationsSize)
        return citationsSize;
    else
    {
        for(int i = citationsSize - 1; i >= 0; i--)
        {
            if(citations[i] >= citationsSize - i)    //如果被引用的次数大于同样次数以上的篇数时
                n++;
            else
                break;
        }
    }
    return n;
}

• 二分查找

int hIndex(int* citations, int citationsSize) {
    int n = citationsSize;
    int low = 0, high = citationsSize - 1;
     while(low <= high)
     {
        int mid  = low + (high-low) / 2;
        if(citations[mid] == n - mid)
            return n-mid;
        else if(citations[mid] < n - mid)
            low = mid + 1;
        else
            high = mid - 1;
    }
    return n-low;
}

• n – mid：代表引用次数大于citations[mid]的篇数
• 如果citations[mid] 小于 篇数，则说明至少被引用citations[mid]次的文章数目大于该最小引用的次数