168. Excel Sheet Column Title

Description

Given an integer columnNumber, return its corresponding column title as it appears in an Excel sheet.

For example:

A -> 1
B -> 2
C -> 3
...
Z -> 26
AA -> 27
AB -> 28 
...

 

Example 1:

Input: columnNumber = 1
Output: "A"

Example 2:

Input: columnNumber = 28
Output: "AB"

Example 3:

Input: columnNumber = 701
Output: "ZY"

 

Constraints:

  • 1 <= columnNumber <= 231 - 1

Solutions

Solution 1

Python Code
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class Solution:
    def convertToTitle(self, columnNumber: int) -> str:
        res = []
        while columnNumber:
            columnNumber -= 1
            res.append(chr(ord('A') + columnNumber % 26))
            columnNumber //= 26
        return ''.join(res[::-1])

Java Code
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class Solution {
    public String convertToTitle(int columnNumber) {
        StringBuilder res = new StringBuilder();
        while (columnNumber != 0) {
            --columnNumber;
            res.append((char) ('A' + columnNumber % 26));
            columnNumber /= 26;
        }
        return res.reverse().toString();
    }
}

Go Code
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func convertToTitle(columnNumber int) string {
	res := []rune{}
	for columnNumber != 0 {
		columnNumber -= 1
		res = append([]rune{rune(columnNumber%26 + int('A'))}, res...)
		columnNumber /= 26
	}
	return string(res)
}

TypeScript Code
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function convertToTitle(columnNumber: number): string {
    let res: string[] = [];
    while (columnNumber > 0) {
        --columnNumber;
        let num: number = columnNumber % 26;
        res.unshift(String.fromCharCode(num + 65));
        columnNumber = Math.floor(columnNumber / 26);
    }
    return res.join('');
}

Rust Code
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impl Solution {
    #[allow(dead_code)]
    pub fn convert_to_title(column_number: i32) -> String {
        let mut ret = String::from("");
        let mut column_number = column_number;

        while column_number > 0 {
            if column_number <= 26 {
                ret.push((('A' as u8) + (column_number as u8) - 1) as char);
                break;
            } else {
                let mut left = column_number % 26;
                left = if left == 0 { 26 } else { left };
                ret.push((('A' as u8) + (left as u8) - 1) as char);
                column_number = (column_number - 1) / 26;
            }
        }

        ret.chars().rev().collect()
    }
}

C# Code
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public class Solution {
    public string ConvertToTitle(int columnNumber) {
        StringBuilder res = new StringBuilder();
        while (columnNumber != 0) {
            --columnNumber;
            res.Append((char) ('A' + columnNumber % 26));
            columnNumber /= 26;
        }
        return new string(res.ToString().Reverse().ToArray());
    }
}