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.
Solution:
- My idea is to sort the array in descending order; find the index c[i] < i+ 1
iterate to compare the index and value
if the value < index i+1, then stop and return the index as the final answer
if value == index i, return the index i+1
else it is the length of the array
1 if len(citations) == 0 or citations is None: 2 return 0 3 citations = sorted(citations, reverse=True) 4 #print ('cita: ', citations) 5 for i, c in enumerate(citations): 6 if i+1 > c: 7 return i 8 elif i+1 == c: 9 return i+1 10 return len(citations)
time complexity: o(nlogn), space complexity o(1)
2 nd method is use hashmap
time complexity o(n), space complexity o(n)
get the array value count, when the value > len(array) L, count into the hashmap[L]
--reference
http://www.cnblogs.com/yrbbest/p/5031910.html
1 if len(citations) == 0 or citations is None: 2 return 0 3 dicCount = {} 4 for ci in citations: 5 if ci > len(citations): 6 if len(citations) not in dicCount: 7 dicCount[len(citations)] = 1 8 else: 9 dicCount[len(citations)] += 1 10 else: 11 if ci not in dicCount: 12 dicCount[ci] = 1 13 else: 14 dicCount[ci] += 1 15 #print (' dicCount : ', dicCount) 16 sum = 0 17 for i in range(len(citations), -1, -1): 18 if i in dicCount: 19 sum += dicCount[i] 20 if sum >= i: 21 return i