【LeetCode从零单刷】H-index I & II

题目：

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 thanh 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.
解答：

一开始做的非常麻烦，遍历引用数（并没连续分布，有间隔值），因此还要计算 maximum、维护与 size 的关系。

遍历应针对连续值（此题中的 size 值是连续分布的）。其实看维基可以清楚知道算法：h-index
(f) = 


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

sort(citations.begin(), citations.end(), [](int a, int b){return a>b;});

vector<int> tmp;
for(int i = 0; i< size; i++)
{
tmp.push_back(citations[i]>(i+1)?(i+1):citations[i]);
}
sort(tmp.begin(), tmp.end(), [](int a, int b){return a>b;});

return tmp[0];
}
};

H-index II 题目有着进一步要求，变为如下：

Follow up for H-Index: What if the 

citations

 array
is sorted in ascending order? Could you optimize your algorithm?

Hint:
Expected runtime complexity is in O(log n) and the input is sorted.
解答：
看到 O（log n）应该自动想到二分搜索。而且这里是已经排好序的 citations 数组。注意处理数组为0，以及数组最大值小于等于0的情况。

class Solution {
public:
int hIndex(vector<int>& citations) {
int size = citations.size();
if(size == 0 || citations[size - 1] <= 0) return 0;

int left = 0;
int right = size - 1;
int mid;
while(left < right)
{
mid = (left + right) / 2;
if (citations[mid] < size - mid)
left = mid + 1;
else
right = mid;
}
return (size - right);
}
};