Given a log of stock prices compute the maximum possible earning.
A zero-indexed array A consisting of N integers is given.
It contains daily prices of a stock share for a period of N consecutive days.
If a single share was bought on day P and sold on day Q, where 0 ≤ P ≤ Q < N,
then the profit of such transaction is equal to A[Q] − A[P], provided that A[Q] ≥ A[P].
Otherwise, the transaction brings loss of A[P] − A[Q].
For example, consider the following array A consisting of six elements such that:
A[0] = 23171
A[1] = 21011
A[2] = 21123
A[3] = 21366
A[4] = 21013
A[5] = 21367
If a share was bought on day 0 and sold on day 2,
a loss of 2048 would occur because A[2] − A[0] = 21123 − 23171 = −2048.
If a share was bought on day 4 and sold on day 5,
a profit of 354 would occur because A[5] − A[4] = 21367 − 21013 = 354.
Maximum possible profit was 356. It would occur if a share was bought on day 1 and sold on day 5.
Write a function,
def solution(A)
that, given a zero-indexed array A consisting of N integers
containing daily prices of a stock share for a period of N consecutive days,
returns the maximum possible profit from one transaction during this period.
The function should return 0 if it was impossible to gain any profit.
For example, given array A consisting of six elements such that:
A[0] = 23171
A[1] = 21011
A[2] = 21123
A[3] = 21366
A[4] = 21013
A[5] = 21367
the function should return 356, as explained above.
Assume that:
N is an integer within the range [0..400,000];
each element of array A is an integer within the range [0..200,000].
方法一:用最小價格取最大利潤
Correctness:100%、Performance:100%
```python def solution(S): minPrice = sys.maxsize profit = 0 for number in A: minPrice = min(number, minPrice) profit = max(number - minPrice, profit) return profit ```
完整練習題 source code 請參閱:github
沒有留言:
張貼留言