Leetcode H-index

Given an array of citations (each citation was a non-negative integer) of a researcher, write a function to compute the RES Earcher ' s h-index.

According to the definition of H-index on Wikipedia: ' A scientist has index H if h of his/her N Papers has at least H citations each, and the other N? h Papers has no more than C7>h citations each. "

For example, given citations = [3, 0, 6, 1, 5] , which means the researcher have papers in total and each of 5 them had received 3, 0, 6, 1, 5 Citatio NS respectively. Since the researcher have 3 papers with at least citations each and the remaining are with 3 no more t Han 3 citations each, he h-index is 3 .

Note: If There is several possible values h for, the maximum one is taken as the h-index.

First sort citations, compare citations[i] and n-i, one is the number of references, and the other is the number of paper that have been quoted so many times at least. H-index should be the smaller of the two. And if citations[i] < N-i, then the next citations[i+1] should be checked to be satisfied with this condition.

1 classsolution (object):2     defHindex (Self, citations):3         """4 : Type Citations:list[int]5 : Rtype:int6         """7         if  notcitations:8             return09 Citations.sort ()Tenn =Len (citations) OneH_index =0 A          forIinchrange (n): -cur = min (citations[i], N-i) -             ifCur >H_index: theH_index =cur -         returnH_index

Or

1 defHindex (Self, citations):2         """3 : Type Citations:list[int]4 : Rtype:int5         """6         if  notcitations:7             return08 Citations.sort ()9n =Len (citations)TenH_index =0 One          forIinchrange (n): A             ifCitations[i] < n-I: -H_index =Citations[i] -             Else: the                 returnN-I -         returnH_index

If the citations is already sorted out. The complexity of the search can be changed to O (LONGN) using the dichotomy method

1 defHindex (Self, citations):2         """3 : Type Citations:list[int]4 : Rtype:int5         """6         7         if  notcitations:8             return09n =Len (citations)Tenleft =0 Oneright = N-1 A          -          whileLeft <=Right : -Mid = (left + right)//2 the             ifCitations[mid] < n-Mid: -left = mid + 1 -             Else: -right = Mid-1 +          -         returnN-left

L13 If this equals sign is not added, an error will occur for the citations list of length =1.

Leetcode H-index