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Update all solutions' changes in 2025-09-11.
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‎en/1-1000/1-two-sum.md‎

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Original link: [1. Two Sum - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/1-two-sum).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).
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‎en/1-1000/13-roman-to-integer.md‎

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# 13. Roman to Integer - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions
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Visit original link: [13. Roman to Integer - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/13-roman-to-integer) for a better experience!
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LeetCode link: [13. Roman to Integer](https://leetcode.com/problems/roman-to-integer), difficulty: **Easy**.
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## LeetCode description of "13. Roman to Integer"
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Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
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```
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Symbol Value
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I 1
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V 5
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X 10
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L 50
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C 100
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D 500
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M 1000
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```
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For example, `2` is written as `II` in Roman numeral, just two ones added together. `12` is written as `XII`, which is simply `X` + `II`. The number `27` is written as `XXVII`, which is `XX` + `V` + `II`.
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Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
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`I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
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`X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
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`C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
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Given a roman numeral, convert it to an integer.
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### [Example 1]
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**Input**: `s = "III"`
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**Output**: `3`
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### [Example 2]
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**Input**: `s = "IV"`
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**Output**: `4`
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### [Example 3]
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**Input**: `s = "IX"`
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**Output**: `9`
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### [Example 4]
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**Input**: `s = "LVIII"`
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**Output**: `58`
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**Explanation**: `L = 50, V= 5, III = 3.`
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### [Example 5]
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**Input**: `s = "MCMXCIV"`
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**Output**: `1994`
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**Explanation**: `M = 1000, CM = 900, XC = 90, IV = 4.`
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### [Constraints]
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- `1 <= s.length <= 15`
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- `s` contains only the characters `('I', 'V', 'X', 'L', 'C', 'D', 'M')`.
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- It is **guaranteed** that `s` is a valid roman numeral in the range `[1, 3999]`.
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### [Hints]
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<details>
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<summary>Hint 1</summary>
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Problem is simpler to solve by working the string from back to front and using a map.
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</details>
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## Intuition
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* The correspondence between characters and values can be represented with a `Map`.
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* The intuition is to add the value to `result` whenever a digit is encountered.
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* But cases like `IV` may occur, so you need to consider whether to process from left to right or from right to left. Which processing direction do you choose?
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<details><summary>Click to view the answer</summary><p> Processing from right to left is more convenient, because once you see that the current character and the previous character form a specific combination, you can handle it directly. </p></details>
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* How do you deal with cases like `IV`?
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<details><summary>Click to view the answer</summary><p> Just handle it in reverse. In the forward direction you do addition, but now you do subtraction. </p></details>
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## Complexity
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- Time complexity: `O(N)`.
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- Space complexity: `O(1)`.
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## Ruby
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```ruby
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# @param {String} s
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# @return {Integer}
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def roman_to_int(s)
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symbol_to_value = {
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'I' => 1,
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'V' => 5,
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'X' => 10,
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'L' => 50,
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'C' => 100,
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'D' => 500,
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'M' => 1000,
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}
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result = 0
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previous_char = nil
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(s.size - 1).downto(0).each do |i|
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char = s[i]
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if ('I' == char && ['V', 'X'].include?(previous_char)) ||
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('X' == char && ['L', 'C'].include?(previous_char)) ||
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('C' == char && ['D', 'M'].include?(previous_char))
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result -= symbol_to_value[char]
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else
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result += symbol_to_value[char]
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end
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previous_char = char
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end
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result
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end
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```
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## Python
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```python
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class Solution:
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def romanToInt(self, s: str) -> int:
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symbol_to_value = {
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'I': 1,
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'V': 5,
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'X': 10,
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'L': 50,
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'C': 100,
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'D': 500,
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'M': 1000,
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}
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result = 0
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previous_char = None
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for i in range(len(s) - 1, -1, -1):
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char = s[i]
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if ('I' == char and previous_char in ['V', 'X']) or \
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('X' == char and previous_char in ['L', 'C']) or \
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('C' == char and previous_char in ['D', 'M']):
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result -= symbol_to_value[char]
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else:
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result += symbol_to_value[char]
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previous_char = char
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return result
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```
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## Other languages
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```java
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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Dear LeetCoders! For a better LeetCode problem-solving experience, please visit website [LeetCode.to](https://leetcode.to): Dare to claim the best practices of LeetCode solutions! Will save you a lot of time!
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Original link: [13. Roman to Integer - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/13-roman-to-integer).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).

‎en/1-1000/15-3sum.md‎

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Original link: [15. 3Sum - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/15-3sum).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).
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‎en/1-1000/151-reverse-words-in-a-string.md‎

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Original link: [151. Reverse Words in a String - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/151-reverse-words-in-a-string).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).
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‎en/1-1000/160-intersection-of-two-linked-lists.md‎

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Original link: [160. Intersection of Two Linked Lists - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/160-intersection-of-two-linked-lists).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).
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‎en/1-1000/169-majority-element.md‎

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# 169. Majority Element - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions
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Visit original link: [169. Majority Element - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/169-majority-element) for a better experience!
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LeetCode link: [169. Majority Element](https://leetcode.com/problems/majority-element), difficulty: **Easy**.
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## LeetCode description of "169. Majority Element"
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Given an array `nums` of size `n`, return _the majority element_.
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The majority element is the element that appears more than `⌊n / 2⌋` times. You may assume that the majority element always exists in the array.
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### [Example 1]
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**Input**: `nums = [3,2,3]`
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**Output**: `3`
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### [Example 2]
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**Input**: `nums = [2,2,1,1,1,2,2]`
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**Output**: `2`
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### [Constraints]
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- `n == nums.length`
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- `1 <= n <= 5 * 10^4`
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- `-10^9 <= nums[i] <= 10^9`
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**Follow-up**: Could you solve the problem in linear time and in `O(1)` space?
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### [Hints]
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<details>
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<summary>Hint 1</summary>
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How to solve the problem in `O(1)` space?
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Please search `Boyer-Moore majority vote algorithm`.
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</details>
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## Intuition
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The key to solving this problem is to use a hash table to store the occurrence count of each `num`. The `key` is the `num`, and the `value` is the number of times it appears.
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## Complexity
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- Time complexity: `O(N)`.
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- Space complexity: `O(N)`.
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## Python
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```python
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class Solution:
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def majorityElement(self, nums: List[int]) -> int:
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num_to_count = defaultdict(int)
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for num in nums:
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num_to_count[num] += 1
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if num_to_count[num] >= len(nums) / 2:
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return num
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```
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## Ruby
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```ruby
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# @param {Integer[]} nums
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# @return {Integer}
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def majority_element(nums)
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num_to_count = Hash.new(0)
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nums.each do |num|
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num_to_count[num] += 1
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if num_to_count[num] > nums.size / 2
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return num
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end
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end
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end
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```
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## Other languages
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```java
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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Dear LeetCoders! For a better LeetCode problem-solving experience, please visit website [LeetCode.to](https://leetcode.to): Dare to claim the best practices of LeetCode solutions! Will save you a lot of time!
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Original link: [169. Majority Element - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/169-majority-element).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).

‎en/1-1000/18-4sum.md‎

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Original link: [18. 4Sum - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/18-4sum).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).
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‎en/1-1000/19-remove-nth-node-from-end-of-list.md‎

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Original link: [19. Remove Nth Node From End of List - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/19-remove-nth-node-from-end-of-list).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).
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‎en/1-1000/20-valid-parentheses.md‎

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Original link: [20. Valid Parentheses - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/20-valid-parentheses).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).
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‎en/1-1000/202-happy-number.md‎

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Original link: [202. Happy Number - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/202-happy-number).
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GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).
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