If you've read the Wikipedia article of H-Index, there is already a neat formula there for computing the h-index, which is written below using the notations of the problem. Note that in the formula below, citations is sorted in descending order and i is 1-indexed.
h = max_i(min(i, citations[i]))
Now you will easily write down the following code.
1 class Solution { 2 public: 3 int hIndex(vector<int>& citations) { 4 sort(citations.rbegin(), citations.rend()); 5 int h = 0, i = 0; 6 for (int c : citations) 7 h = max(h, min(++i, c)); 8 return h; 9 } 10 };
This code takes 20ms. In fact, rbegin and rend seems to be relatively slow. An alternative is to sort citations in normal ascending order and then count all those papers with citations larger than their indexes, as what Stefan does here. Now the code runs in 12ms.
1 class Solution { 2 public: 3 int hIndex(vector<int>& citations) { 4 sort(citations.begin(), citations.end()); 5 int h = 0, i = citations.size(); 6 for (int c : citations) 7 h += (c > --i); 8 return h; 9 } 10 };
Well, both the above codes are in O(nlogn) time. Is there a linear time solution? The answer is yes: refer to this post if you like :-)
1 class Solution { 2 public: 3 int hIndex(vector<int>& citations) { 4 int n = citations.size(), h = 0; 5 int* counts = new int[n + 1](); 6 for (int c : citations) 7 counts[min(c, n)]++; 8 for (int i = n; i; i--) { 9 h += counts[i]; 10 if (h >= i) return i; 11 } 12 return h; 13 } 14 };
This code uses both linear time and space, and runs in 8ms. Wow, we've moved a long way from the original 20ms version :-)
