628. Maximum Product of Three Numbers

本文探讨了如何在整数数组中找到三个数,使它们的乘积达到最大,并提供了两种解决方案:一种是通过冒泡排序找到最大的三个数和最小的两个数;另一种则是直接使用数组排序的方法。

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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.

========================================================================

Ideas: 

Arrays.sort()  can solve this problem.But quick sorting is not a good way to do it. We can use bubble sorting to find the largest three and the smallest two numbers,The result is largest in product of the largest three numbers and the maximum of the product of the smallest two numbers and the maximum number.

========================================================================

Code:

class Solution {
    public int maximumProduct(int[] nums) {
        if (nums.length==3){
            return nums[0]*nums[1]*nums[2];
        }
        for (int i = 0; i < 3; i++) {
            int min = nums[i];
            int point = i;
            for (int j = i; j < nums.length-i-1; j++) {
                if (nums[j]>nums[j+1]){
                    exchange(nums,j,j+1);
                }
                if (min>nums[j]){
                    min = nums[j];
                    point = j;
                }
            }
            exchange(nums,i,point);
        }
        int last = nums.length-1;
        return Math.max(nums[1]*nums[0]*nums[last],nums[last]*nums[last-1]*nums[last-2]);
    }
    private void exchange(int[] arr,int index1,int index2){
        int temp = arr[index1];
        arr[index1] = arr[index2];
        arr[index2] = temp;
    }
}

or

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

 

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