628. 三个数的最大乘积

https://leetcode-cn.com/problems/maximum-product-of-three-numbers/

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-07-25 12:26:56
 * @Description: https://leetcode-cn.com/problems/maximum-product-of-three-numbers/
 * @FilePath: \leetcode-googtech\#628. Maximum Product of Three Numbers\Solution.java
 */ 
class Solution {
    public int maximumProduct(int[] nums) {
        // 初始化第一,第二以及第三大元素
        int firstMax = Integer.MIN_VALUE, secondMax = Integer.MIN_VALUE,thirdMax = Integer.MIN_VALUE;
        // 初始化第一,第二以及第三小元素
        int firstMin = Integer.MAX_VALUE, secondMin = Integer.MAX_VALUE;
        // 遍历数组
        for(int num : nums) {
            // 获取第一,第二以及第三大元素
           if(num > firstMax) {
                thirdMax = secondMax;
                secondMax = firstMax;
                firstMax = num;
           }else if(num > secondMax) {
               thirdMax = secondMax;
               secondMax = num;
           }else if(num > thirdMax) {
               thirdMax = num;
           }
           // 获取第一,第二小元素
           if(num < firstMin) {
               secondMin = firstMin;
               firstMin = num;
           }else if(num < secondMin) {
               secondMin = num;
           }
        }
       // 有正有负有零: 最大3个数相乘 / 最小2个数(负负得正)与最大的1个数相乘
       return Math.max(firstMax * secondMax * thirdMax,firstMax * firstMin * secondMin);
    }
}

Python

'''
@Author: Goog Tech
@Date: 2020-07-25 12:27:03
@Description: https://leetcode-cn.com/problems/maximum-product-of-three-numbers/
@FilePath: \leetcode-googtech\#628. Maximum Product of Three Numbers\Solution.py
'''
class Solution(object):
    def maximumProduct(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # 初始化第一、第二、第三大元素以及第一、第二小元素
        # -sys.maxsize = -9223372036854775807
        firstMax = secondMax = thirdMax = -sys.maxsize
        firstMin = secondMin = sys.maxsize
        # 遍历数组
        for num in nums:
            # 获取第一,第二以及第三大元素
            if num > firstMax:
                thirdMax, secondMax, firstMax = secondMax, firstMax, num
            elif num > secondMax:
                thirdMax, secondMax = secondMax, num
            elif num > thirdMax:
                thirdMax = num
            # 获取第一,第二小元素
            if num < firstMin:
                secondMin, firstMin = firstMin, num
            elif num < secondMin:
                secondMin = num
        # 有正有负有零: 最大3个数相乘 / 最小2个数(负负得正)与最大的1个数相乘
        return max(firstMax * secondMax * thirdMax,
                   firstMax * firstMin * secondMin)