628. Maximum Product of Three Numbers(Java)_maximumproductofthreenumbers java-优快云博客

本文介绍了一种算法问题，即如何从整数数组中找出三个数使其乘积最大，并给出了两种解决方案：一种是通过排序实现，时间复杂度为O(nlogn)；另一种是在O(n)时间内直接找到最大的三个数和最小的两个数。

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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.

代码1：先排序，时间复杂度O(nlogn) ，空间复杂度O(1)

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

代码2：时间复杂度O(n)，空间复杂度O(1)

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