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

Note:

1. The length of the given array will be in range [3,10^4] 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.

一个数组中3个数的最大乘积，可能存在的两种情况，一种是这个乘积来自最大的三个数，另一种是最大的一个数乘以最小的两个数。（可以自己仔细考虑一下其中的正负情况）

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

以上解法中的排序部分其实并不是必须的，因为我们需要知道的只是最大的三个数和最小的两个数

public class Solution {
    public int maximumProduct(int[] nums) {
        int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
        int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;
        for (int n: nums) {
            if (n <= min1) {
                min2 = min1;
                min1 = n;
            } else if (n <= min2) {     // n lies between min1 and min2
                min2 = n;
            }
            if (n >= max1) {            // n is greater than max1, max2 and max3
                max3 = max2;
                max2 = max1;
                max1 = n;
            } else if (n >= max2) {     // n lies betweeen max1 and max2
                max3 = max2;
                max2 = n;
            } else if (n >= max3) {     // n lies betwen max2 and max3
                max3 = n;
            }
        }
        return Math.max(min1 * min2 * max1, max1 * max2 * max3);
    }
}