Given a positive integer num, split it into two non-negative integers num1 and num2 such that:
num1 and num2 is a permutation of num.\n\n\tnum1 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:
\n\nnum does not contain any leading zeros.num1 and num2 may differ from the order of occurrence of num.\n
Example 1:
\n\n\nInput: num = 4325\nOutput: 59\nExplanation: We can split 4325 so that\n\nnum1is 24 andnum2is 35, giving a sum of 59. We can prove that 59 is indeed the minimal possible sum.\n
Example 2:
\n\n\nInput: num = 687\nOutput: 75\nExplanation: We can split 687 so that\n\nnum1is 68 andnum2is 7, which would give an optimal sum of 75.\n
\n
Constraints:
\n\n10 <= num <= 109给你一个正整数 num ，请你将它分割成两个非负整数 num1 和 num2 ，满足：
num1 和 num2 直接连起来，得到 num 各数位的一个排列。\n\n\tnum1 和 num2 中所有数字出现的次数之和等于 num 中所有数字出现的次数。num1 和 num2 可以包含前导 0 。请你返回 num1 和 num2 可以得到的和的 最小 值。
注意：
\n\nnum 保证没有前导 0 。num1 和 num2 中数位顺序可以与 num 中数位顺序不同。\n\n
示例 1：
\n\n\n输入：num = 4325\n输出：59\n解释：我们可以将 4325 分割成\n\nnum1= 24 和num2= 35 ，和为 59 ，59 是最小和。\n
示例 2：
\n\n\n输入：num = 687\n输出：75\n解释：我们可以将 687 分割成\n\nnum1= 68 和num2= 7 ，和为最优值 75 。\n
\n\n
提示：
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