[LeetCode]274 H指数

H-Index(H指数)

【难度：Medium】
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.
给定记录了一位学者论文被引用次数的一个数组，求该名学者的H指数。
H指数是指一名科学家在已发表的N篇论文中，有h篇被引用次数不少于h次的论文，剩余的N-h篇论文被引用次数不超过h次。

---

解题思路

理解h指数的含义，先对数组升序排序后，找到临界值即为H指数。
关键点在与判断当前论文被引用的次数是否不小于剩下的论文数量，由于升序排列，如果不小于说明之后的论文被引用次数都满足条件，H指数就找到了。

---

c++代码如下：

class Solution {
public:
    int hIndex(vector<int>& citations) {
        if (citations.empty())
            return 0;
        int len = citations.size();
        quicksort(citations,0,len-1);

        for (int i = 0; i < len; i++) {
            if (citations[i] >= (len-i)) {
                return len-i;
            }
        }
        return 0;
    }
    void quicksort(vector<int>& t, int low, int high) {
        if (low >= high)
            return;
        int key = t[low];
        int i = low;
        int j = high;
        while (i < j) {
            while (i < j && t[j] >= key)
                j--;
            t[i] = t[j];
            while (i < j && t[i] <= key)
                i++;
            t[j] = t[i];

        }
        t[i] = key;

        quicksort(t,low,j-1);
        quicksort(t,j+1,high);
    }
};