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.
一名科研人员的h指数是指他至多有h篇论文分别被引用了至少h次。
用hash[i]表示引用i次的文章数。如果i>n，用hash[n]计数。
从后往前遍历hash数组，sum累计hash[i]，找到第一个sum>=i的i返回。
class Solution {
public:
int hIndex(vector<int>& citations) {
int n = citations.size();
vector<int> hash(n+1,0);
for (int i = 0; i < n; i++){
if (citations[i]>=n){
hash[n] += 1;
}
else{
hash[citations[i]] += 1;
}
}
int sum = 0;
for (int i = n; i >= 0; i--){
sum += hash[i];
if (sum >= i){
return i;
}
}
return 0;
}
};citations array is sorted in ascending order? Could you optimize your algorithm?class Solution {
public:
int hIndex(vector<int>& citations) {
int left = 0, right = citations.size() - 1;
while (left <= right){
int mid = left + (right - left) / 2;
if (citations.size() - mid == citations[mid]){
return citations.size()-mid;
}
else if (citations[mid] > citations.size() - mid){
right = mid - 1;
}
else{
left = mid + 1;
}
}
return citations.size()-left;//
}
};
