codelity--MaxProductOfThree

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