Given a positive integer?num, split it into two non-negative integers?num1?and?num2?such that:

• The concatenation of?num1?and?num2?is a permutation of?num.

• In other words, the sum of the number of occurrences of each digit in?num1?and?num2?is equal to the number of occurrences of that digit in?num.

• num1?and?num2?can contain leading zeros.

Return?the?minimum?possible sum of?num1?and?num2.

Notes:

• It is guaranteed that?num?does not contain any leading zeros.

• The order of occurrence of the digits in?num1?and?num2?may differ from the order of occurrence of?num.

?

Example 1:

Input: num = 4325

Output: 59

Explanation: We can split 4325 so that num1 is 24 and num2 is 35, giving a sum of 59. We can prove that 59 is indeed the minimal possible sum.

Example 2:

Input: num = 687

Output: 75

Explanation: We can split 687 so that num1 is 68 and num2 is 7, which would give an optimal sum of 75.

?

Constraints:

• 10 <= num <= 109

Easy 題目：

?把數(shù)據(jù)放到數(shù)組中之后，排序，然后根據(jù)數(shù)組的長度是奇數(shù)還是偶數(shù)去累加數(shù)字即可。

奇數(shù)我寫了一個函數(shù)，偶數(shù)也寫了一個函數(shù)。然后就通過了。

Runtime:?1 ms, faster than?50.00%?of?Java?online submissions for?Split With Minimum Sum.

Memory Usage:?39.3 MB, less than?50.00%?of?Java?online submissions for?Split With Minimum Sum.