Count digits in a number
Easy
Java
Control Flow
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Problem Statement
Task: return how many digits are in an integer. Handle 0 properly.
Input Format
An integer n.
Output Format
An integer: number of digits in n.
Constraints
• n can be 0, positive, or negative
• Must handle 0 correctly
• Must handle 0 correctly
Concept Explanation
Digits are counted by dividing the number by 10 repeatedly.
Special case: 0 has exactly one digit.
Special case: 0 has exactly one digit.
Step-by-Step Explanation
1.Read integer n.
2.If n == 0, return 1.
3.If n is negative, make it positive.
4.Initialize count = 0.
5.While n > 0:
• Increment count.
• Divide n by 10.
6.After loop ends, return count.
2.If n == 0, return 1.
3.If n is negative, make it positive.
4.Initialize count = 0.
5.While n > 0:
• Increment count.
• Divide n by 10.
6.After loop ends, return count.
Concept Explanation
Digits are counted by dividing the number by 10 repeatedly.
Special case: 0 has exactly one digit.
Special case: 0 has exactly one digit.
Step-by-Step Explanation
1.Read integer n.
2.If n == 0, return 1.
3.If n is negative, make it positive.
4.Initialize count = 0.
5.While n > 0:
• Increment count.
• Divide n by 10.
6.After loop ends, return count.
2.If n == 0, return 1.
3.If n is negative, make it positive.
4.Initialize count = 0.
5.While n > 0:
• Increment count.
• Divide n by 10.
6.After loop ends, return count.
Input / Output Format
Input Format
An integer n.
Output Format
An integer: number of digits in n.
Constraints
• n can be 0, positive, or negative
• Must handle 0 correctly
• Must handle 0 correctly
Examples
Input:
0
Output:
1
Example Solution (Public)
static int countDigits(int x){if(x==0) return 1;int n=x<0?-x:x;int c=0;while(n>0){c++;n/=10;}return c;}
Official Solution Code
static int countDigits(int x){if(x==0) return 1;int n=x<0?-x:x;int c=0;while(n>0){c++;n/=10;}return c;}
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