628. Maximum Product of Three Numbers

最新推荐文章于 2020-01-23 14:07:22 发布

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最新推荐文章于 2020-01-23 14:07:22 发布

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分类专栏： leetcode

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本文介绍了一种高效算法，用于从整数数组中找出三个数的最大乘积。通过一次遍历实现，考虑了正数与负数组合的可能性。

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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：最大值可能是3个正数相乘或者1正两个最小的负数相乘。

class Solution {
public:
    int maximumProduct(vector<int>& nums) {
        int pMax, psMax, ptMax;
        pMax = psMax = ptMax = INT_MIN;
        int nMin, nsMin;
        nMin = nsMin = INT_MAX;
        for(int i = 0; i < nums.size(); i++){
            int t = nums[i];
            if(t > pMax) {
                ptMax = psMax;
                psMax = pMax;
                pMax = t;                
            } else if(t > psMax) {
                ptMax = psMax;
                psMax = t;                
            } else if(t > ptMax) 
                ptMax = t;
            if(t < nMin){
                nsMin = nMin;
                nMin = t;
            } else if(t < nsMin)
                nsMin = t;        
        }        
        return max(pMax * psMax * ptMax, pMax * nMin * nsMin);
    }
};