Maximum Product of Three Numbers问题解法
2017-07-23 09:19
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问题描述:
Given an integer array, find three numbers whose product is maximum and output the maximum product.
示例:
问题分析:
本题求解一个数组中三个元素的积的最大值,可先将数组升序排序,得到会出现最大值的元素组合first、sencod、third、fouth,通过比较得出答案。
过程详见代码:
class Solution {
public:
int maximumProduct(vector<int>& nums) {
int first, second, third,fouth;
sort(nums.begin(), nums.end());
first = nums[nums.size() - 1] * nums[nums.size() - 2] * nums[nums.size() - 3];
second = nums[0] * nums[1] * nums[2];
third = nums[0] * nums[1] * nums[nums.size() - 1];
fouth = nums[0] * nums[nums.size() - 2] * nums[nums.size() - 1];
return max(max(first,second),max(third,fouth));
}
};
Given an integer array, find three numbers whose product is maximum and output the maximum product.
示例:
Input: [1,2,3] Output: 6
Input: [1,2,3,4] Output: 24
问题分析:
本题求解一个数组中三个元素的积的最大值,可先将数组升序排序,得到会出现最大值的元素组合first、sencod、third、fouth,通过比较得出答案。
过程详见代码:
class Solution {
public:
int maximumProduct(vector<int>& nums) {
int first, second, third,fouth;
sort(nums.begin(), nums.end());
first = nums[nums.size() - 1] * nums[nums.size() - 2] * nums[nums.size() - 3];
second = nums[0] * nums[1] * nums[2];
third = nums[0] * nums[1] * nums[nums.size() - 1];
fouth = nums[0] * nums[nums.size() - 2] * nums[nums.size() - 1];
return max(max(first,second),max(third,fouth));
}
};
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