leetcode-628-easy()-其他

leetcode-628-easy()

Maximum Product of Three Numbers

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

Example 1:

Input: nums = [1,2,3]
Output: 6
Example 2:

Input: nums = [1,2,3,4]
Output: 24
Example 3:

Input: nums = [-1,-2,-3]
Output: -6
Constraints:

3 <= nums.length <= 104
-1000 <= nums[i] <= 1000

思路一：对给定的数组排序后完全分类，数组只可能出现三种情况

• 在零轴中间
• 全部位于零轴左边
• 全部位于零轴右边

所以只要判断数组左右乘积最大值即可

public int maximumProduct(int[] nums) {
    Arrays.sort(nums);

    int max = nums[0] * nums[1] * nums[nums.length - 1];
    max = Math.max(nums[nums.length - 1] * nums[nums.length - 2] * nums[nums.length - 3], max);

    return max;
}

思路二：对思路一可以进行优化，不用对数组进行全部排序，只需要取最大的三个值和最小的三个值判断

Maximum Product of Three Numbers

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

Example 1:

Input: nums = [1,2,3]
Output: 6
Example 2:

Input: nums = [1,2,3,4]
Output: 24
Example 3:

Input: nums = [-1,-2,-3]
Output: -6
Constraints:

3 <= nums.length <= 104
-1000 <= nums[i] <= 1000

思路一：对给定的数组排序后完全分类，数组只可能出现三种情况

• 在零轴中间
• 全部位于零轴左边
• 全部位于零轴右边

所以只要判断数组左右乘积最大值即可

public int maximumProduct(int[] nums) {
    Arrays.sort(nums);

    int max = nums[0] * nums[1] * nums[nums.length - 1];
    max = Math.max(nums[nums.length - 1] * nums[nums.length - 2] * nums[nums.length - 3], max);

    return max;
}

思路二：对思路一可以进行优化，不用对数组进行全部排序，只需要取最大的三个值和最小的三个值判断