1689. Partitioning Into Minimum Number Of Deci-Binary Numbers (M)

Partitioning Into Minimum Number Of Deci-Binary Numbers (M)

题目

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 <= 10^5
• n consists of only digits.
• n does not contain any leading zeros and represents a positive integer.

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题意

一个 十-二 数指只包含1和0且不以0开头的数，给定一个任意整数，判断最少需要多少个 十-二 数相加能得到这个整数。

思路

直觉上就是求字符串中最大的数字，直接AC。

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代码实现

Java

class Solution {
    public int minPartitions(String n) {
        int ans = 0;
        for (char c : n.toCharArray()) {
            if (c - '0' > ans) {
                ans = c - '0';
            }
        }
        return ans;
    }
}