LeetCode 0136. Single Number Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

Description

Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

You must implement a solution with a linear runtime complexity and use only constant extra space.

 

Example 1:

Input: nums = [2,2,1]

Output: 1

Example 2:

Input: nums = [4,1,2,1,2]

Output: 4

Example 3:

Input: nums = [1]

Output: 1

 

Constraints:

• 1 <= nums.length <= 3 * 104
• -3 * 104 <= nums[i] <= 3 * 104
• Each element in the array appears twice except for one element which appears only once.

Solutions

Solution 1: Bitwise Operation

The XOR operation has the following properties:

• Any number XOR 0 is still the original number, i.e., x \oplus 0 = x;
• Any number XOR itself is 0, i.e., x \oplus x = 0;

Performing XOR operation on all elements in the array will result in the number that only appears once.

The time complexity is O(n), where n is the length of the array. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustJavaScriptC#CSwift

class Solution: def singleNumber(self, nums: List[int]) -> int: return reduce(xor, nums)(code-box)

class Solution { public int singleNumber(int[] nums) { int ans = 0; for (int v : nums) { ans ^= v; } return ans; } }(code-box)

class Solution { public: int singleNumber(vector<int>& nums) { int ans = 0; for (int v : nums) { ans ^= v; } return ans; } };(code-box)

func singleNumber(nums []int) (ans int) { for _, v := range nums { ans ^= v } return }(code-box)

function singleNumber(nums: number[]): number { return nums.reduce((r, v) => r ^ v); }(code-box)

impl Solution { pub fn single_number(nums: Vec<i32>) -> i32 { nums.into_iter().reduce(|r, v| r ^ v).unwrap() } }(code-box)

/** * @param {number[]} nums * @return {number} */ var singleNumber = function (nums) { return nums.reduce((a, b) => a ^ b); };(code-box)

public class Solution { public int SingleNumber(int[] nums) { return nums.Aggregate(0, (a, b) => a ^ b); } }(code-box)

int singleNumber(int* nums, int numsSize) { int ans = 0; for (int i = 0; i < numsSize; i++) { ans ^= nums[i]; } return ans; }(code-box)

class Solution { func singleNumber(_ nums: [Int]) -> Int { return nums.reduce(0, ^) } }(code-box)

Solution 2

Java

class Solution { public int singleNumber(int[] nums) { return Arrays.stream(nums).reduce(0, (a, b) -> a ^ b); } }(code-box)