leecode 解题总结：274. H-Index

#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <map>
#include <unordered_map>
#include <algorithm>
using namespace std;
/*
问题:
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.

分析:高被引次数为h表示至少有h篇文章引用次数都>=h，其余N-h篇文章引用次数<=h。就是以h作为统计
的分割点。那么其实取值返回就是[0,最高引用次数]
比如[0,0,0]，那么h=0
比如[5,5,5,5,5],那么h=5
如果采用暴力破解，初始选择引用次数最多的作为h，统计h出现的次数，如果次数< h,则h--。

输入:
5(数组元素个数）
3 0 6 1 5
1
100
输出:
3
1

Input:[100]
Output:0
Expected:1
出错的原因在于：尝试的引用次数不是递减的，而是map中的键
高被引次数最多不会超过文章个数，

关键:leecode提示
先排序，计算可能的值，使用额外空间

参考解法：http://www.cnblogs.com/zslhq/p/5954582.html
1、排序，从后向前遍历，对于当前元素i，存储到集合中，此时引用次数为nums[i]，
篇数为n-i,判断其是否满足至少有h篇文章引用次数>=h
引用次数>=h
 nums[i](引用次数) >= n-i(篇数)
举例:3 0 6 1 5,排序后: 0 1 3 5 6

2、不排序的方法就是采用统计排序，先计算最大值，然后统计，
统计后从后向前仍然采用上述方式
至少有h篇引用次数>=h(篇数)，即引用次数>=篇数
篇数可以统计，引用次数为当前值
*/

class Solution {
public:

    int hIndex2(vector<int>& citations) {
		if(citations.empty())
		{
			return 0;
		}
		sort(citations.begin() , citations.end());
		int size = citations.size();
		int result = 0;
		for(int i = size - 1 ; i >= 0 ; i--)
		{
			//满足至少有引用次数>=h(篇数)，记录结果
			if( citations.at(i) >= size - i )
			{
				result = size - i;
			}
			else
			{
				break;
			}
		}
		return result;
    }
};

void process()
{
	 vector<int> nums;
	 int value;
	 int num;
	 Solution solution;
	 vector<int> result;
	 while(cin >> num )
	 {
		 nums.clear();
		 for(int i = 0 ; i < num ; i++)
		 {
			 cin >> value;
			 nums.push_back(value);
		 }
		 int answer = solution.hIndex(nums);
		 cout << answer << endl;
	 }
}

int main(int argc , char* argv[])
{
	process();
	getchar();
	return 0;
}