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:

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

Example 2:

  1. Input: n = "82734"
  2. Output: 8

Example 3:

  1. Input: n = "27346209830709182346"
  2. 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.

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。

代码实现

Java

  1. class Solution {
  2.     public int minPartitions(String n) {
  3.         int ans = 0;
  4.         for (char c : n.toCharArray()) {
  5.             if (c - '0' > ans) {
  6.                 ans = c - '0';
  7.             }
  8.         }
  9.         return ans;
  10.     }
  11. }