H index

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本文介绍了一种计算研究者h指数的方法，h指数反映了一位科学家的论文引用情况。通过排序和遍历论文引用数量，或者利用二分查找在已排序的引用数组中找到满足h指数条件的值。

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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."

Example:

Input: citations = [3,0,6,1,5]
Output: 3 
Explanation: [3,0,6,1,5] 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, her h-index is 3.

/* h-index 计算 作者有h篇引用数量在h以上的文章
 * 按引用数量排序 遍历到第i篇
 * 从小到大 排序  当前引用量>n-i时
 * */
class Solution {
public:
    int hIndex(vector<int>& citations) {
        sort(citations.begin(), citations.end());
        int len=citations.size();
        for(int i=0;i<len;i++){
            if(citations[i]>=len-i) return len-i;
        }
        return 0;
    }
};

如果是有序的：

/* h-index 计算 作者有h篇引用数量在h以上的文章
 * 现在数组已经按升序排序完毕 求h index
 * 有序数组上查找 首先想到二分
 * N-i为第为满足h index的最小引用数  不够的话 i右移动  否则左移动
 * */
class Solution {
public:
    int hIndex(vector<int>& citations) {
        int l=0, r=citations.size()-1, n=citations.size();
        while(l<=r){
            int mid = (l+r)/2;
            if(citations[mid]==n-mid) return n-mid;
            else if(citations[mid]<n-mid)   l = mid +1;// 引用不够 考虑更高的引用数
            else    r = mid - 1;
        }
        return n-l;
    }
};