【leetcode】1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

题目如下：

A decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary, while 112 and 3001 are not.

Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n.

Example 1:

Input: n = "32"
Output: 3
Explanation: 10 + 11 + 11 = 32

Example 2:

Input: n = "82734"
Output: 8

Example 3:

Input: n = "27346209830709182346"
Output: 9

Constraints:

• 1 <= n.length <= 105
• n consists of only digits.
• n does not contain any leading zeros and represents a positive integer.

解题思路：本题很有意思，仔细想想，其实就是求出n中所有位中的最大值，即为结果。

代码如下：

class Solution(object):
    def minPartitions(self, n):
        """
        :type n: str
        :rtype: int
        """
        res = 0
        max_num = -float('inf')
        for i in str(n):
            if max_num == -float('inf'):
                max_num = int(i)
                res = int(i)
            elif int(i) > max_num:
                res += int(i) - max_num
                max_num = int(i)
        return res