Leetcode 75

Solution/136. Single Number/136. Single Number.py

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+class Solution:
+  def singleNumber(self, nums: list[int]) -> int:
+    return functools.reduce(operator.xor, nums, 0)

Solution/136. Single Number/readme.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/0100-0199/0136.Single%20Number/README_EN.md
+tags:
+    - Bit Manipulation
+    - Array
+---
+
+<!-- problem:start -->
+
+# [136. Single Number](https://leetcode.com/problems/single-number)
+
+[中文文档](/solution/0100-0199/0136.Single%20Number/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>Given a <strong>non-empty</strong>&nbsp;array of integers <code>nums</code>, every element appears <em>twice</em> except for one. Find that single one.</p>
+
+<p>You must&nbsp;implement a solution with a linear runtime complexity and use&nbsp;only constant&nbsp;extra space.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [2,2,1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [4,1,2,1,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
+	<li><code>-3 * 10<sup>4</sup> &lt;= nums[i] &lt;= 3 * 10<sup>4</sup></code></li>
+	<li>Each element in the array appears twice except for one element which appears only once.</li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### 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 $O(1)$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def singleNumber(self, nums: List[int]) -> int:
+        return reduce(xor, nums)
+```
+
+#### Java
+
+```java
+class Solution {
+    public int singleNumber(int[] nums) {
+        int ans = 0;
+        for (int v : nums) {
+            ans ^= v;
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int singleNumber(vector<int>& nums) {
+        int ans = 0;
+        for (int v : nums) {
+            ans ^= v;
+        }
+        return ans;
+    }
+};
+```
+
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->