1689. Partitioning Into Minimum Number Of Deci Binary Numbers - kumaeki/LeetCode GitHub Wiki
1689. Partitioning Into Minimum Number Of Deci-Binary Numbers
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.
解法1
每个数字位上每一次只能加1,或者不加(加0), 所以各个位置上最大的数就是最后的答案.
class Solution {
public int minPartitions(String n) {
char res = '1';
for(char c : n.toCharArray()){
if(c > res)
res = c;
}
return (int)(res - '0');
}
}