Maximum Product of Three Numbers

问题：

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.

解决：

① 要考虑到值为负数的情况！

public class Solution {
public int maximumProduct(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
Arrays.sort(nums);
int a = nums[nums.length - 1] * nums[nums.length - 2] * nums[nums.length - 3];
int b = nums[0] * nums[1] * nums[nums.length - 1];
return a > b ? a : b;
}
}

② 线性遍历数组，依次找到最大的三个值和最小的两个值，因为三个负数相乘为负数。

public class Solution {
public int maximumProduct(int[] nums) {
int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;
for (int n: nums) {
if (n <= min1) {
min2 = min1;
min1 = n;
} else if (n <= min2) {
min2 = n;
}
if (n >= max1) {
max3 = max2;
max2 = max1;
max1 = n;
} else if (n >= max2) {
max3 = max2;
max2 = n;
} else if (n >= max3) {
max3 = n;
}
}
return Math.max(min1 * min2 * max1, max1 * max2 * max3);
}
}