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.

Note: If there are several possible values for h, the maximum one is taken as the h-index.

题目大意：

　　给定一位研究者论文被引用次数的数组，输出研究者的 h 指数。h指数指他的 （N 篇论文中）至多有 h 篇论文分别被引用了至少 h 次。

理  解：

　　首先对数组从大到小排序，使用二分法进行比较。

　　若 当前引用篇数 <= 当前最小引用数量，则向右继续找。保存当前指数=当前引用篇数，更新最大h指数。

　　若 当前引用篇数 > 当前最小引用数量，则向左找。

代  码 C++：

　　

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int n = citations.size();
        if(n<=0) return 0;
        if(n==1 && citations[0]>=1) 
        {
            return 1;
        }
        int i,j,temp;
        for(i=0;i<n;++i){
            for(j=i;j<n;++j){
                if(citations[i]<citations[j]){
                    temp = citations[i];
                    citations[i] = citations[j];
                    citations[j] = temp;
                }
            }
        }
        if(n<=citations[n-1])return n;
        int left=0,right=n-1,mid;
        int h = 0; 
        while(left<right){
            mid = (left+right)/2;
            if(mid+1<=citations[mid]){
                if(mid+1>h) h = mid+1;
                left = mid+1;
            }else{
                right = mid;
            }
        }
        return h;
    }
};