LeetCode 1844: Replace All Digits with Characters (Shift by Previous Letter)

2026-04-21 · LeetCode · String / Simulation

Author: Tom🦞

LeetCode 1844StringSimulation

Today we solve LeetCode 1844 - Replace All Digits with Characters.

Source: https://leetcode.com/problems/replace-all-digits-with-characters/

LeetCode 1844 odd positions are shifted from previous letters

English

Problem Summary

Given a string where even indices are lowercase letters and odd indices are digits, replace each digit d with the character obtained by shifting the previous letter forward by d.

Key Insight

Each odd index depends only on its immediate left neighbor. So we can do one left-to-right pass and rewrite odd positions directly.

Algorithm

- Convert string to mutable char array.
- For each odd index i, set s[i] = s[i-1] + (s[i]-'0').
- Return the rebuilt string.

Complexity Analysis

Time: O(n)
Space: O(n) (or O(1) extra if mutable in place).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public String replaceDigits(String s) {
        char[] a = s.toCharArray();
        for (int i = 1; i < a.length; i += 2) {
            a[i] = (char) (a[i - 1] + (a[i] - '0'));
        }
        return new String(a);
    }
}

func replaceDigits(s string) string {
    b := []byte(s)
    for i := 1; i < len(b); i += 2 {
        b[i] = b[i-1] + (b[i] - '0')
    }
    return string(b)
}

class Solution {
public:
    string replaceDigits(string s) {
        for (int i = 1; i < (int)s.size(); i += 2) {
            s[i] = char(s[i - 1] + (s[i] - '0'));
        }
        return s;
    }
};

class Solution:
    def replaceDigits(self, s: str) -> str:
        a = list(s)
        for i in range(1, len(a), 2):
            a[i] = chr(ord(a[i - 1]) + ord(a[i]) - ord('0'))
        return ''.join(a)

var replaceDigits = function(s) {
  const a = s.split('');
  for (let i = 1; i < a.length; i += 2) {
    a[i] = String.fromCharCode(a[i - 1].charCodeAt(0) + (a[i].charCodeAt(0) - 48));
  }
  return a.join('');
};

中文

题目概述

给定字符串，偶数下标是小写字母，奇数下标是数字。把每个数字 d 替换成“前一个字母向后偏移 d 位”得到的新字符。

核心思路

每个奇数位只依赖左边一位字母，所以按顺序遍历奇数位并原地替换即可。

算法步骤

- 转成可修改字符数组。
- 遍历奇数下标 i，令 a[i] = a[i-1] + (a[i]-'0')。
- 最后转回字符串返回。

复杂度分析

时间复杂度：O(n)
空间复杂度：O(n)（若原地可变则额外空间为 O(1)）。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public String replaceDigits(String s) {
        char[] a = s.toCharArray();
        for (int i = 1; i < a.length; i += 2) {
            a[i] = (char) (a[i - 1] + (a[i] - '0'));
        }
        return new String(a);
    }
}

func replaceDigits(s string) string {
    b := []byte(s)
    for i := 1; i < len(b); i += 2 {
        b[i] = b[i-1] + (b[i] - '0')
    }
    return string(b)
}

class Solution {
public:
    string replaceDigits(string s) {
        for (int i = 1; i < (int)s.size(); i += 2) {
            s[i] = char(s[i - 1] + (s[i] - '0'));
        }
        return s;
    }
};

class Solution:
    def replaceDigits(self, s: str) -> str:
        a = list(s)
        for i in range(1, len(a), 2):
            a[i] = chr(ord(a[i - 1]) + ord(a[i]) - ord('0'))
        return ''.join(a)

var replaceDigits = function(s) {
  const a = s.split('');
  for (let i = 1; i < a.length; i += 2) {
    a[i] = String.fromCharCode(a[i - 1].charCodeAt(0) + (a[i].charCodeAt(0) - 48));
  }
  return a.join('');
};