Middle-题目42/43：274. H-Index && 275. H-Index II

题目原文：
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.

$275. H-Index II$

Follow up for H-Index: What if the citations array is sorted in ascending order? Could you optimize your algorithm?
题目大意：
$274. H-Index$

给出一个研究者的引用量数组，写一个算法计算这个研究者的h指数。
H指数的定义：一个人的h指数是指在一定期间内他发表的论文至少有h篇的被引频次不低于h次。
例如，给出citations数组=[3,0,6,1,5],代表这个研究者写了5篇论文，引用次数分别是3,0,6,1,5.那么他的h值是3,因为有3篇论文的引用次数>=3次。

$275. H-Index II$

若citations数组是升序排列的，能否优化算法？
题目分析：
根据wiki中给出的算法：

把引用数降序排列画在坐标系中，位于直线y=x上面的点个数就是h值。由于本题中引用数是降序排列的，所以改为统计位于直线x+y=n(n为论文数目)上方点的数目即可。因为数组是有序的，所以可以使用二分查找。
源码：（language：java）

public class Solution {
    public int hIndex(int[] citations) {
        int count=0,length = citations.length;
        int start=0,end=length-1,mid=0;
        while(start<=end) {
            mid = (start+end) / 2;
            if(citations[mid] >= length-mid) {
                if(mid==0 || citations[mid-1]<length-mid+1)
                    return length-mid;
                else if(citations[mid-1]>=length-mid)
                    end=mid-1;
            }
            else
                start=mid+1;
        }
        return 0;       
    }
}

成绩：
H-Index:4ms,beats 9.28%,众数1ms,38.82%
H-Index II:13ms,beats 31.68%,众数12ms,24.72%
Cmershen的碎碎念：
在第一题中，使用二分查找比无脑查找慢，但第二题中无脑插的耗时是213ms，远差于二分查找。