LeetCode 2160. Minimum Sum of Four Digit Number After Splitting Digits

You are given a positive integer num consisting of exactly four digits. Split num into two new integers new1 and new2 by using the digits found in num. Leading zeros are allowed in new1 and new2, and all the digits found in num must be used.

• For example, given num = 2932, you have the following digits: two 2's, one 9 and one 3. Some of the possible pairs [new1, new2] are [22, 93], [23, 92], [223, 9] and [2, 329].

Return the minimum possible sum of new1 and new2.

Example 1:

Input: num = 2932
Output: 52
Explanation: Some possible pairs [new1, new2] are [29, 23], [223, 9], etc.
The minimum sum can be obtained by the pair [29, 23]: 29 + 23 = 52.

Example 2:

Input: num = 4009
Output: 13
Explanation: Some possible pairs [new1, new2] are [0, 49], [490, 0], etc. 
The minimum sum can be obtained by the pair [4, 9]: 4 + 9 = 13.

Constraints:

• 1000 <= num <= 9999

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这题已经给定了这个数字长为4，把每个数字拆开来重新组合一下，求最小的组合。那就直接把这个数字里面的数字拆开排序，两个最小的当十位，两个最大的当个位就行。

class Solution {
    public int minimumSum(int num) {
        int[] array = new int[4];
        int count = 0;
        while (num != 0) {
            array[count] = num % 10;
            num /= 10;
            count++;
        }
        Arrays.sort(array);
        return (array[0] + array[1]) * 10 + array[2] + array[3];
    }
}