## 题目描述

Given an integer array, find three numbers whose product is maximum and output the maximum product.

Example 1:

``````Input: [1,2,3]
Output: 6``````

Example 2:

``````Input: [1,2,3,4]
Output: 24``````

Note:

1. The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].
2. Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer.

## 解法一

### Python

``````class Solution(object):
def maximumProduct(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
max_1 = -sys.maxsize
max_2 = -sys.maxsize
max_3 = -sys.maxsize
min_1 = sys.maxsize
min_2 = sys.maxsize
for num in nums:
if num > max_1:
max_3 = max_2
max_2 = max_1
max_1 = num
elif num > max_2:
max_3 = max_2
max_2 = num
elif num > max_3:
max_3 = num
if num < min_1:
min_2 = min_1
min_1 = num
elif num < min_2:
min_2 = num
return max(max_1 * max_2 * max_3, max_1 * min_1 * min_2)``````

### Java

``````public class Solution {
public int maximumProduct(int[] nums) {
int max1 = Integer.MIN_VALUE;
int max2 = Integer.MIN_VALUE;
int max3 = Integer.MIN_VALUE;
int min1 = Integer.MAX_VALUE;
int min2 = Integer.MAX_VALUE;
for (int num : nums) {
if (num > max1) {
max3 = max2;
max2 = max1;
max1 = num;
} else if (num > max2) {
max3 = max2;
max2 = num;
} else if (num > max3) {
max3 = num;
}
if (num < min1) {
min2 = min1;
min1 = num;
} else if (num < min2) {
min2 = num;
}
}
return Math.max(max1 * max2 * max3, max1 * min1 * min2);
}
}``````

### C++

``````class Solution {
public:
int maximumProduct(vector<int> &nums) {
int max1 = INT_MIN;
int max2 = INT_MIN;
int max3 = INT_MIN;
int min1 = INT_MAX;
int min2 = INT_MAX;
for (int num : nums) {
if (num > max1) {
max3 = max2;
max2 = max1;
max1 = num;
} else if (num > max2) {
max3 = max2;
max2 = num;
} else if (num > max3) {
max3 = num;
}
if (num < min1) {
min2 = min1;
min1 = num;
} else if (num < min2) {
min2 = num;
}
}
return max(max1 * max2 * max3, max1 * min1 * min2);
}
};``````