Description
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.
Solution
O(N^3)的暴力破解會超時，這道題的核心思想其實是已知要求3個數(shù)的乘積最大，那么要么是數(shù)組里面三個最大的正數(shù)之積咽筋，要么就是最小的兩個負數(shù)乘上最大的正數(shù)。
Sort & Choose：time O(N * lgN), space O(1)
public class Solution {
public int maximumProduct(int[] nums) {
int len = nums.length;
Arrays.sort(nums);
int max = Math.max(nums[len - 3] * nums[len - 2] * nums[len - 1]
, nums[0] * nums[1] * nums[len - 1]);
return max;
}
}
Iteration & Choose：time O(N * lgN), space O(1)
因為不需要數(shù)組完全有序碍脏，所以我們只需要記錄最大的三個值和最小的兩個值就可以了。
public class Solution {
public int maximumProduct(int[] nums) {
int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;
int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
for (int n : nums) {
if (n > max1) {
max3 = max2;
max2 = max1;
max1 = n;
} else if (n > max2) {
max3 = max2;
max2 = n;
} else if (n > max3) {
max3 = n;
}
// don't merge decisions below with decisions above!
if (n < min1) {
min2 = min1;
min1 = n;
} else if (n < min2) {
min2 = n;
}
}
return Math.max(max1 * max2 * max3, min1 * min2 * max1);
}
}
