Easy
You are given a positive integer n.
Return the maximum product of any two digits in n.
Note: You may use the same digit twice if it appears more than once in n.
Example 1:
Input: n = 31
Output: 3
Explanation:
n are [3, 1].3 * 1 = 3.Example 2:
Input: n = 22
Output: 4
Explanation:
n are [2, 2].2 * 2 = 4.Example 3:
Input: n = 124
Output: 8
Explanation:
n are [1, 2, 4].1 * 2 = 2, 1 * 4 = 4, 2 * 4 = 8.Constraints:
10 <= n <= 109public class Solution {
public int maxProduct(int n) {
int m1 = n % 10;
n /= 10;
int m2 = n % 10;
n /= 10;
while (n > 0) {
int a = n % 10;
if (a > m1) {
if (m1 > m2) {
m2 = m1;
}
m1 = a;
} else {
if (a > m2) {
m2 = a;
}
}
n /= 10;
}
return m1 * m2;
}
}