思路
一开始只考虑排序,有可能两个小的负数之积更大,一次线性扫描,找两个最小的和三个最大的即可
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 (int n : nums) {
if (n < min1) {
min2 = min1;
min1 = n;
}
else if (n < min2) {
min2 = n;
}
if (n > max3) {
max1 = max2;
max2 = max3;
max3 = n;
}
else if (n > max2) {
max1 = max2;
max2 = n;
}
else if (n > max1) {
max1 = n;
}
}
return max(min1 * min2 * max3, max1 * max2 * max3);
}
};