給定一個整型數組，在數組中找出由三個數組成的最大乘積，并輸出這個乘積。
- 示例 1:
輸入: [1,2,3]
輸出: 6
- 示例 2:
輸入: [1,2,3,4]
輸出: 24
- 注意:
給定的整型數組長度范圍是[3,104]，數組中所有的元素范圍是[-1000, 1000]。
輸入的數組中任意三個數的乘積不會超出32位有符號整數的范圍。
- 先排序
class Solution {
public:void quickSort(vector<int>& nums, int low, int high) {if (high <= low) {return;}int key = nums[low];int i = low;int j = high+1;while(true) {while(nums[++i] < key) {if (i == high) {break;}}while(nums[--j] > key) {if (j == low) {break;}}if (i >=j ) {break;}int temp = nums[i];nums[i] = nums[j];nums[j] = temp;}nums[low] = nums[j];nums[j] = key;quickSort(nums, low, j-1);quickSort(nums, j+1, high);}int maximumProduct(vector<int>& nums) {quickSort(nums, 0, nums.size()-1);return max(nums[nums.size()-1] * nums[nums.size()-2] * nums[nums.size()-3], nums[nums.size()-1] * nums[0] * nums[1]);}
};
- 線性掃描
class Solution {
public:int maximumProduct(vector<int>& nums) {int min1 = INT_MAX, min2 = INT_MAX;int max1 = INT_MIN, max2 = INT_MIN, max3 = INT_MIN;for (const auto& num: nums) {if (num < min1) {min2 = min1;min1 = num;} else if (num < min2) {min2 = num;}if (num > max1) {max3 = max2;max2 = max1;max1 = num;} else if (num > max2) {max3 = max2;max2 = num;} else if (num > max3){max3 = num;}}return max(max1 * max2 * max3, max1 * min1 * min2);}
};
來源：力扣（LeetCode）
