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# 628.Maximum-Product-of-Three-Numbers

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

## 代码

### Approach 1: Brute Force

Time complexity : O(n^3)

### Approach 2: Using Sorting

public class Solution {
public int maximumProduct(int[] nums) {
Arrays.sort(nums);
// it could also be possible that two negative numbers lying at the left extreme
return Math.max(nums[0] * nums[1] * nums[nums.length - 1], nums[nums.length - 1] * nums[nums.length - 2] * nums[nums.length - 3]);
}
}

### Approach #3 Single Scan

Complexity Analysis
• Time complexity : O(n). Only one iteration over the nums array of length n is required.
• Space complexity : O(1). Constant extra space is used.
class Solution {
public int maximumProduct(int[] nums) {
int min1 = Integer.MAX_VALUE;
int min2 = Integer.MAX_VALUE;
int max1 = Integer.MIN_VALUE;
int max2 = Integer.MIN_VALUE;
int max3 = Integer.MIN_VALUE;
for (int n: nums) {
if (n <= min1) {
min2 = min1;
min1 = n;
} else if (n <= min2) {
min2 = n;
}
if (n >= max1) {
max3 = max2;
max2 = max1;
max1 = n;
} else if (n >= max2) {
max3 = max2;
max2 = n;
} else if (n >= max3) {
max3 = n;
}
}
return Math.max(min1 * min2 * max1, max1 * max2 * max3);
}
}