问题描述
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.
问题分析
简单做法：将数组排序后，遍历数组找到citation>=n-i的n-i即为h，算法复杂度即排序的复杂度O(nlgn)；
高效做法：设置一个记录数组count，遍历一遍citations记录count，由count得到结果，算法复杂度O(n)。
AC代码
#encoding=utf-8
class Solution(object):
def hIndex(self, citations):
"""
:type citations: List[int]
:rtype: int
"""
# 注释掉的是朴素做法
# n = len(citations)
# if n == 0:
# return 0
# citations.sort()
# for i in range(n):
# paper = n-i
# if citations[i] >= paper:
# return paper
# return 0
n = len(citations)
if n == 0:
return 0
count = [0 for i in range(n+1)]
for ci in citations:
if ci > n:
count[n] += 1
else:
count[ci] += 1
accumu = 0
for i in range(n, -1, -1):
accumu += count[i]
if accumu >= i:
return i
return 0
