Description

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.

Solution

O(N^3)的暴力破解会超时，这道题的核心思想其实是已知要求3个数的乘积最大，那么要么是数组里面三个最大的正数之积，要么就是最小的两个负数乘上最大的正数。

Sort & Choose：time O(N * lgN), space O(1)

public class Solution {
    public int maximumProduct(int[] nums) {
        int len = nums.length;
        Arrays.sort(nums);
        int max = Math.max(nums[len - 3] * nums[len - 2] * nums[len - 1]
                          , nums[0] * nums[1] * nums[len - 1]);
        return max;
    }
}

Iteration & Choose：time O(N * lgN), space O(1)

因为不需要数组完全有序，所以我们只需要记录最大的三个值和最小的两个值就可以了。

public class Solution {
    public int maximumProduct(int[] nums) {
        int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;
        int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
        
        for (int n : nums) {
            if (n > max1) {
                max3 = max2;
                max2 = max1;
                max1 = n;
            } else if (n > max2) {
                max3 = max2;
                max2 = n;
            } else if (n > max3) {
                max3 = n;
            } 
            
            // don't merge decisions below with decisions above!
            if (n < min1) {
                min2 = min1;
                min1 = n;
            } else if (n < min2) {
                min2 = n;
            }
        }
        
        return Math.max(max1 * max2 * max3, min1 * min2 * max1);
    }
}