628. Maximum Product of Three Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/u013527178/article/details/79724073

本文介绍了一种算法，用于从整数数组中找到三个数，使得这三个数的乘积最大，并提供了三种不同的实现方式。文章还讨论了考虑正负数情况下的最优解策略。

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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:

The length of the given array will be in range [3,10⁴] and all elements are in the range [-1000, 1000].
Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer.

思路：

1.排序后进行选择；

2.遍历。考虑到

1）三个数都是非负数，那这三个数都是正数中最大的；

2）三个都是负数，那这三个数都是负数中最大的；

3）两负一正，两负一定是负数中最小的，正数是最大的；

归纳为，找三个最大的和两个最小的，进行积的比较

代码1:

class Solution {
public:
    int maximumProduct(vector<int>& nums) {
        vector<int>::iterator max1,max2,max3,min1,min2,min3;
        int vmax1,vmax2,vmax3,vmin1,vmin2,vmin3;
        int result=0;
        max1=max_element(nums.begin(),nums.end());
        vmax1=*max1;
        nums.erase(max1);
        
        max2=max_element(nums.begin(),nums.end());
        vmax2=*max2;
        nums.erase(max2);
        
        min1=min_element(nums.begin(),nums.end());
        vmin1=*min1;
        nums.erase(min1);
        // 3 numbers
        if(nums.size()==0) {
            result = (vmax1)*(vmin1)*(vmax2);
            return result;
        }
        // 4 numbers
        else if(nums.size()==1){
            int temp1 = (vmax1)*(nums[0])*(vmax2);
            int temp2 = (nums[0])*(vmin1)*(vmax1);
            return temp1>temp2?temp1:temp2;
        }
        // 5 numbers
        else if(nums.size()==2){
            min2=min_element(nums.begin(),nums.end());
            vmin2=*min2;
            nums.erase(min2);
            int temp1 = (vmax1)*(nums[0])*(vmax2);
            int temp2 = (vmax1)*(vmin1)*(vmin2);
            return temp1>temp2?temp1:temp2;
        }
        else{
            min2=min_element(nums.begin(),nums.end());
            vmin2=*min2;
            nums.erase(min2);
            max3=max_element(nums.begin(),nums.end());
            vmax3=*max3;
            nums.erase(max3);
            int temp1 = (vmax1)*(vmax2)*(vmax3);
            int temp2 = (vmax1)*(vmin1)*(vmin2);
            return temp1>temp2?temp1:temp2;
        }
        
    }
};

代码2：

class Solution {
public:
    int maximumProduct(vector<int>& nums) {
        auto max1=max_element(nums.begin(),nums.end());
        int vmax1=*max1;
        nums.erase(max1);
        
        auto max2=max_element(nums.begin(),nums.end());
        int vmax2=*max2;
        nums.erase(max2);
        
        auto max3=max_element(nums.begin(),nums.end());
        int vmax3=*max3;
        nums.erase(max3);
        
        int res;
        if(nums.size()==0){res=vmax1*vmax2*vmax3;}
        else if(nums.size()==1) {vmax1*vmax2*vmax3>vmax1*vmax2*nums[0]?res=vmax1*vmax2*vmax3:res=vmax1*vmax2*nums[0];}
        else{
            auto min1=min_element(nums.begin(),nums.end());
            int vmin1=*min1;
            nums.erase(min1);
            auto min2=min_element(nums.begin(),nums.end());
            int vmin2=*min2;
            nums.erase(min2);
            vmax1*vmax2*vmax3>vmax1*vmin2*vmin1?res=vmax1*vmax2*vmax3:res=vmax1*vmin2*vmin1;
        }
        return res;
    }
};

代码3：

int maximumProduct(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    int n = nums.size();
    int temp1 = nums[n-1]*nums[n-2]*nums[n-3];
    int temp2 = nums[0]*nums[1]*nums[n-1];
    return temp1>temp2?temp1:temp2;
}