274. 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.

s思路:
1. 涨指数了，h-index这么计算的。首先想想简单粗暴的计算：排序。例如,[3, 0, 6, 1, 5]递减排序得到[6,5,3,1,0],然后遍历，如果遍历到3,3等于坐标+1，所以继续遍历，1就小于坐标+1，所以返回此时的坐标值。
2. 这样的复杂度就是o(nlgn)。能做到o(n)吗？
3. 看了提示。有o(n)。考虑到h值的边界是0-citations.size();也就是说，h值的范围是confined within size；也就是说，我们可以先扫描一遍，凡是值大于citations.size()的都可以用citations.size()替换，而不影响计算结果；也就是说，现在所有的值也是confined的，妥妥用counting sort就够了，因此复杂度就降低到o(n)。太TM妙了！
4. 找到输出的边界，一切就清楚了！

//方法1：class Solution {public:    int hIndex(vector<int>& citations) {        //        sort(citations.begin(),citations.end(),[](int&a,int&b){             return a>b;           });        for(int i=0;i<citations.size();i++){            if(citations[i]<i+1) return i;            }        return citations.size();    }};//方法2：先对数据做“削顶处理”，先对超出范围的数削顶，然后就是counting sort，美！class Solution {public:    int hIndex(vector<int>& citations) {        //        int n=citations.size();        vector<int> count(n+1,0);        for(int i=0;i<citations.size();i++){            if(citations[i]>n){                count[n]++;            }else                count[citations[i]]++;          }        int sum=0;        for(int i=n;i>=0;i--){            sum+=count[i];            if(sum>=i) return i;            }        return n;    }};