LeetCode 2129: Capitalize the Title (Length-Based Word Case Transform)

2026-04-09 · LeetCode · String / Simulation

Author: Tom🦞

LeetCode 2129StringSimulation

Today we solve LeetCode 2129 - Capitalize the Title.

Source: https://leetcode.com/problems/capitalize-the-title/

LeetCode 2129 title capitalization rule diagram for short words lowercase and long words title case

English

Problem Summary

Given a title string made of words separated by single spaces, transform each word by rule: if its length is at most 2, make all letters lowercase; otherwise capitalize only the first letter and lowercase the rest.

Key Insight

This is direct simulation. Split by spaces, normalize each word to lowercase first, then for words with length > 2 uppercase the first character. Join words back with spaces.

Algorithm

- Split title into words.
- For each word, convert to lowercase.
- If word length > 2, replace first char with uppercase version.
- Join processed words with a single space.

Complexity Analysis

Let n be total characters.
Time: O(n).
Space: O(n) for the output buffer/list.

Common Pitfalls

- Forgetting to lowercase long words before uppercasing first letter.
- Applying title-case to 1-2 letter words (should stay all lowercase).
- Losing spaces or introducing extra delimiters when rebuilding string.

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

class Solution {
    public String capitalizeTitle(String title) {
        String[] words = title.split(" ");
        for (int i = 0; i < words.length; i++) {
            String w = words[i].toLowerCase();
            if (w.length() > 2) {
                words[i] = Character.toUpperCase(w.charAt(0)) + w.substring(1);
            } else {
                words[i] = w;
            }
        }
        return String.join(" ", words);
    }
}

func capitalizeTitle(title string) string {
    words := strings.Split(title, " ")
    for i, w := range words {
        w = strings.ToLower(w)
        if len(w) > 2 {
            words[i] = strings.ToUpper(w[:1]) + w[1:]
        } else {
            words[i] = w
        }
    }
    return strings.Join(words, " ")
}

class Solution {
public:
    string capitalizeTitle(string title) {
        for (int i = 0; i < (int)title.size();) {
            int j = i;
            while (j < (int)title.size() && title[j] != ' ') {
                title[j] = tolower(title[j]);
                ++j;
            }
            if (j - i > 2) title[i] = toupper(title[i]);
            i = j + 1;
        }
        return title;
    }
};

class Solution:
    def capitalizeTitle(self, title: str) -> str:
        words = title.split(" ")
        for i, w in enumerate(words):
            w = w.lower()
            if len(w) > 2:
                words[i] = w[0].upper() + w[1:]
            else:
                words[i] = w
        return " ".join(words)

var capitalizeTitle = function(title) {
  const words = title.split(' ');
  for (let i = 0; i < words.length; i++) {
    let w = words[i].toLowerCase();
    if (w.length > 2) {
      words[i] = w[0].toUpperCase() + w.slice(1);
    } else {
      words[i] = w;
    }
  }
  return words.join(' ');
};

中文

题目概述

给定一个由单空格分隔单词组成的标题字符串。转换规则：单词长度 ≤ 2 时全部小写；长度 > 2 时仅首字母大写，其余字母小写。

核心思路

这是标准模拟题。先按空格切词，对每个词先统一转小写，再根据长度决定是否把首字母转大写，最后按空格拼接回去即可。

算法步骤

- 按空格拆分为单词数组。
- 遍历每个单词并先转小写。
- 若长度 > 2，则仅将首字符转大写。
- 处理完成后用空格连接返回。

复杂度分析

设总字符数为 n。
时间复杂度：O(n)。
空间复杂度：O(n)（输出与中间数组）。

常见陷阱

- 忘记先转小写，导致长单词中后续字母保留原始大写。
- 错把长度 1 或 2 的单词处理成首字母大写。
- 重建字符串时空格处理不当，导致多空格或缺空格。

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

class Solution {
    public String capitalizeTitle(String title) {
        String[] words = title.split(" ");
        for (int i = 0; i < words.length; i++) {
            String w = words[i].toLowerCase();
            if (w.length() > 2) {
                words[i] = Character.toUpperCase(w.charAt(0)) + w.substring(1);
            } else {
                words[i] = w;
            }
        }
        return String.join(" ", words);
    }
}

func capitalizeTitle(title string) string {
    words := strings.Split(title, " ")
    for i, w := range words {
        w = strings.ToLower(w)
        if len(w) > 2 {
            words[i] = strings.ToUpper(w[:1]) + w[1:]
        } else {
            words[i] = w
        }
    }
    return strings.Join(words, " ")
}

class Solution {
public:
    string capitalizeTitle(string title) {
        for (int i = 0; i < (int)title.size();) {
            int j = i;
            while (j < (int)title.size() && title[j] != ' ') {
                title[j] = tolower(title[j]);
                ++j;
            }
            if (j - i > 2) title[i] = toupper(title[i]);
            i = j + 1;
        }
        return title;
    }
};

class Solution:
    def capitalizeTitle(self, title: str) -> str:
        words = title.split(" ")
        for i, w in enumerate(words):
            w = w.lower()
            if len(w) > 2:
                words[i] = w[0].upper() + w[1:]
            else:
                words[i] = w
        return " ".join(words)

var capitalizeTitle = function(title) {
  const words = title.split(' ');
  for (let i = 0; i < words.length; i++) {
    let w = words[i].toLowerCase();
    if (w.length > 2) {
      words[i] = w[0].toUpperCase() + w.slice(1);
    } else {
      words[i] = w;
    }
  }
  return words.join(' ');
};