标签：nlogn pap solution init res ever else function .so
https://leetcode.com/problems/h-index/#/description
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 hcitations 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 3citations 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.
Sol 1:
class Solution(object): def hIndex(self, citations): """ :type citations: List[int] :rtype: int """ # 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。 This means the index of element has something to do with the value of element/ # Time O(nlogn) Space O(1) citations.sort() self.reverse(citations) for i in range(len(citations)): if i + 1 == citations[i]: return i + 1 if i + 1 > citations[i]: return i return len(citations) def reverse(self, nums): left = 0 right = len(nums) - 1 while left < right: tmp = nums[left] nums[left] = nums[right] nums[right] = tmp left += 1 right -=1 return nums
Sol 2:
class Solution(object): def hIndex(self, citations): """ :type citations: List[int] :rtype: int """ # Time O(n) Space O(n) n = len(citations) citeCount = [0] * (n+1) for c in citations: if c >= n: citeCount[n] += 1 else: citeCount[c] += 1 sum = 0 # current number of papers for i in range(n, -1, -1): sum += citeCount[i] if sum >= i: return i return 0
标签：nlogn pap solution init res ever else function .so
原文地址：http://www.cnblogs.com/prmlab/p/7227978.html
