Here, We see Excel Sheet Column Number LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Excel Sheet Column Number LeetCode Solution

Problem Statement

Given a string columnTitle that represents the column title as appears in an Excel sheet, return its corresponding column number.

For example:A -> 1 B -> 2 C -> 3 … Z -> 26 AA -> 27 AB -> 28 …

Example 1:
Input: columnTitle = “A”
Output: 1

Example 2:
Input: columnTitle = “AB”
Output: 28

Example 3:
Input: columnTitle = “ZY”
Output: 701

Excel Sheet Column Number Leetcode Solution C++

class Solution {
public:
    int titleToNumber(string columnTitle) {
        int result = 0;
        for(char c : columnTitle)
        {
            int d = c - 'A' + 1;
            result = result * 26 + d;
        }
        return result;
    }
};Code language: PHP (php)

Excel Sheet Column Number Leetcode Solution Java

public class Solution {
    public int titleToNumber(String columnTitle) {
        if (columnTitle == null) return -1;
        int sum = 0;
        for (char c : columnTitle.toUpperCase().toCharArray()) {
            sum *= 26;
            sum += c - 'A' + 1;
        }
        return sum;
    }
}Code language: PHP (php)

Excel Sheet Column Number Solution JavaScript

var titleToNumber = function(columnTitle) {
    const charCodeBase = 'A'.charCodeAt(0) - 1;
    const n = columnTitle.length;
    let number = 0;
    for (let i = 0; i < n; i++)
        number += (columnTitle.charCodeAt(i) - charCodeBase) * Math.pow(26, n-i-1);
    return number;
};Code language: JavaScript (javascript)

Excel Sheet Column Number Solution Python

class Solution(object):
    def titleToNumber(self, columnTitle):
            corresponding_number = 0
            for i in range(len(columnTitle)):
                current_letter=columnTitle[i]
                corresponding_letter_value=ord(current_letter) - ord('A') + 1
                corresponding_number += (corresponding_letter_value *pow(26,len(columnTitle)-i-1))
            return corresponding_number