# LeetCode_274: H-index

H指数的计算，这道题是典型的排序思路，博主第一次使用一般的排序方法写，后来发现整数这种用基数排序更合适，同时需要考虑很多特例，博主贴了这两份代码。由于测试案例的原因，基数排序速度上的优势没有体现出来。

### 原题

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.

### 一般的排序解法

class Solution {
public:
int hIndex_old(vector<int>& citations) {
sort(citations.rbegin(), citations.rend());
const int len = citations.size();
if (len == 0){
return 0;
}
for (int i = 0; i < len; ++i){
if (citations[i] < i+1){
return i;
}
}
return len;
}
};

### 基数排序解法

class Solution {
public:
int hIndex(vector<int>& citations){
const int len = citations.size();
if (len == 0){
return 0;
}

vector<int> count(len+1);
for (int i = 0; i < len; ++i){
if (citations[i] >= len){
count[len]++;
}
else if (citations[i] >= 0){
count[citations[i]]++;
}
else{ //ciatations[i] < 0
}
}
int total_cite = 0;
for (int i = len; i >= 0; --i){
total_cite += count[i];
if (total_cite >= i){
return i;
}
}
return 0;
}
};