标签：tps 使用 ref init min 复杂度 type http 长度
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.
先对citations 排序，比较citations[i] 和 n - i， 一个是引用数，一个是至少被引用了这么多次的paper的数量。 h-index应该是二者比较小的那一个。 而且如果citations[i] < n - i , 那么应该检查下一个citations[i+1]是不是满足这个条件。
1 class Solution(object): 2 def hIndex(self, citations): 3 """ 4 :type citations: List[int] 5 :rtype: int 6 """ 7 if not citations: 8 return 0 9 citations.sort() 10 n = len(citations) 11 h_index = 0 12 for i in range(n): 13 cur = min(citations[i], n - i) 14 if cur > h_index: 15 h_index = cur 16 return h_index
或者
1 def hIndex(self, citations): 2 """ 3 :type citations: List[int] 4 :rtype: int 5 """ 6 if not citations: 7 return 0 8 citations.sort() 9 n = len(citations) 10 h_index = 0 11 for i in range(n): 12 if citations[i] < n - i: 13 h_index = citations[i] 14 else: 15 return n - i 16 return h_index
如果citations已经是排序好了的。可以使用二分法使得搜索的复杂度变成O(longN)
1 def hIndex(self, citations): 2 """ 3 :type citations: List[int] 4 :rtype: int 5 """ 6 7 if not citations: 8 return 0 9 n = len(citations) 10 left = 0 11 right = n-1 12 13 while left <= right: 14 mid = (left + right)//2 15 if citations[mid] < n - mid: 16 left = mid + 1 17 else: 18 right = mid - 1 19 20 return n - left
L13如果不加这个等号，对于长度=1的citations list会出错。
标签：tps 使用 ref init min 复杂度 type http 长度
原文地址：http://www.cnblogs.com/lettuan/p/6181810.html
