Sort (2) -- H-Index I, II

H-Index

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.

解法1：使用数组记录每个引用的值有几篇文章

之前的写的没保存，贴一个Java的

public int hIndex(int[] citations) {
	int L = citations.length;
	if(L<1) return 0;
	int[] counts = new int[L+1];
	for(int i : citations) {
		if(i>L) counts[L]++;   //hIndex <= length
		else counts[i]++;
	}
	int res = 0;
	for(int k=L; k>=0; k--) {    //采用累加的方式计算引用次数，这样就不需要每次更新比自己大的引用数
	    res += counts[k];
	    if(res>=k) return k;
	}
	return 0;
}

解法2：利用快排的思想，每次迭代选择一个枢纽点，使得数组的左侧小于枢纽点，右侧大于枢纽点。如果枢纽点的值大于右侧的数，则尝试对左边迭代找更大的h-index；反之，则在右边找满足条件的hIndex.

    int partition(vector<int>& citations, int left, int right){
        int mid = (left + right) / 2;   //pivot的选择也可以是其他的，对于随机数组而言，选第一个也可以
        int pivot = citations[mid];
        swap(citations[mid], citations[right]);
        int i = left - 1, j = right;
        for (;;){
            while (citations[++i] < pivot);
            while (j > 0 && citations[--j] > pivot);   //median3的快排针对长度大于2的数组；这里要加入j的条件防止越界
            if (i < j) swap(citations[i], citations[j]);
            else break;
        }
        swap(citations[i], citations[right]);
        return i;
    }

    int hIndex(vector<int>& citations) {
        int n = citations.size();
        int hIndex = 0;
        int left = 0, right = n - 1;
        while (left <= right){
            int index = partition(citations, left, right);
            if (n - index <= citations[index]){   // n-index篇文章有至少n-index次引用
                hIndex = n-index;
                right = index - 1;   //试图找更大的hIndex
            }
            else left = index + 1;
        }
        return hIndex;
    }

H-Index II

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

    int hIndex(vector<int>& citations) {
        int n = citations.size();
        int left = 0;
        int right = n - 1;
        int hIndex = 0;
        while (left <= right){
            int half = (left + right) / 2;
            if (citations[half] >= n - half){
                hIndex = n - half;
                right = half - 1;
            }
            else left = half + 1;
        }
        return hIndex;
    }