codelity--MaxProductOfThree
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trick:
if using python, we can swap two elements like \
A[k], A[minimal] = A[minimal], A[k]
problem:
A non-empty zero-indexed array A consisting of N integers is given. The product of triplet (P, Q, R) equates to A[P] * A[Q] * A[R] (0 ≤ P < Q < R < N).
For example, array A such that:
A[0] = -3
A[1] = 1
A[2] = 2
A[3] = -2
A[4] = 5
A[5] = 6
contains the following example triplets:
(0, 1, 2), product is −3 * 1 * 2 = −6
(1, 2, 4), product is 1 * 2 * 5 = 10
(2, 4, 5), product is 2 * 5 * 6 = 60
Your goal is to find the maximal product of any triplet.
Write a function:
def solution(A)
that, given a non-empty zero-indexed array A, returns the value of the maximal product of any triplet.
For example, given array A such that:
A[0] = -3
A[1] = 1
A[2] = 2
A[3] = -2
A[4] = 5
A[5] = 6
the function should return 60, as the product of triplet (2, 4, 5) is maximal.
Assume that:
N is an integer within the range [3..100,000];
each element of array A is an integer within the range [−1,000..1,000].
Complexity:
expected worst-case time complexity is O(N*log(N));
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
解答:
需要注意,排序后,最大值不一定是前三个,因为可能会有两个或者以上的很大的负数 例如[-5,-5,4,2]
code:
def solution(A): # write your code in Python 2.7 A.sort(reverse = True) neg_num = 0 for x in A: if x < 0: neg_num += 1 if A[0] < 0: return A[0]*A[1]*A[2] else: if neg_num >= 2: res1 = A[0]*A[1]*A[2] res2 = A[0]*A[-1]*A[-2] if res1 > res2: return res1 else: return res2 else: return A[0]*A[1]*A[2] pass
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