Given a positive integer num, split it into two non-negative integers num1and num2such that:

The concatenation of num1and num2is a permutation of num.

In other words, the sum of the number of occurrences of each digit in num1and num2is equal to the number of occurrences of that digit in num.

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num1and num2can contain leading zeros.

Return the minimum possible sum of num1and num2.

Notes:

It is guaranteed that numdoes not contain any leading zeros.

The order of occurrence of the digits in num1and num2may 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 num2is 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 num1is 68 and num2 is 7, which would give an optimal sum of 75.

Constraints:

10 <= num <= 109

Easy 题目：

把数据放到数组中之后，排序，然后根据数组的长度是奇数还是偶数去累加数字即可。

奇数我写了一个函数，偶数也写了一个函数。然后就通过了。

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.

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