動機
把二分搜最討厭的部分完全展示的一題
Problem
Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in an ascending order, return compute the researcher's h-index.
According to the definition of h-index on Wikipedia: A scientist has an index h if h of their n papers have at least h citations each, and the other n − h papers have no more than h citations each.
If there are several possible values for h, the maximum one is taken as the h-index.
You must write an algorithm that runs in logarithmic time.
Example 1:
Input: citations = [0,1,3,5,6]Output: 3Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had received 0, 1, 3, 5, 6 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, their h-index is 3.
Example 2:
Input: citations = [1,2,100]Output: 2
Constraints:
n == citations.length1 <= n <= 1050 <= citations[i] <= 1000citationsis sorted in ascending order.
Sol
h-index的定義是
前N個paper的引用數大於等於N
因為已經排序了，所以第i個，代表前i個的引用數都小於等於他
所以h-index就是總長減i
但痛苦的是
- i是index
- h-index的range是 0~總長，故i的範圍是0~總長
如果要處理的漂亮要在
- 終止條件
- 算中點
- 縮小範圍時
下功夫
Ver1
- 終止條件: 包含右邊
- 算中點: 無條件進位
- 縮小範圍時: 不保留原本的index，兩邊都往內縮
class Solution:
def hIndex(self, cs: List[int]) -> int:
if not cs:
return 0
i = 0
j = len(cs)-1
while i <= j:
mid = (i+j)//2 if (i+j) % 2 == 0 else (i+j+1)//2 # 無條件進位
print(i,j,mid)
if cs[mid] == len(cs) - mid:
return cs[mid]
elif cs[mid] > len(cs) - mid:
j = mid-1
else:
i = mid+1
print(i,j)
return len(cs)-i
Ver2
- 終止條件: 不包含右邊
- 算中點: 無條件捨去
- 縮小範圍時: 保留右邊的index
class Solution:
def hIndex(self, cs: List[int]) -> int:
if not cs:
return 0
i = 0
j = len(cs)
while i < j:
mid = (i+j)//2 # 無條件捨去
if cs[mid] == len(cs) - mid:
return cs[mid]
elif cs[mid] > len(cs) - mid:
j = mid
else:
i = mid+1
return len(cs)-i