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Java-based LeetCode algorithm problem solutions, regularly updated
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136\. Single Number
Easy
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.
To solve the "Single Number" problem in Java with a `Solution` class, we'll use bitwise XOR operation. Below are the steps:
1. **Create a `Solution` class**: Define a class named `Solution` to encapsulate our solution methods.
2. **Create a `singleNumber` method**: This method takes an array `nums` as input and returns the single number that appears only once.
3. **Initialize a variable to store the result**: Initialize a variable `singleNumber` to 0.
4. **Iterate through the array and perform bitwise XOR operation**: Iterate through the array `nums`. For each number `num` in the array, perform bitwise XOR operation with the `singleNumber`.
5. **Return the result**: After iterating through the entire array, the `singleNumber` variable will store the single number that appears only once. Return `singleNumber`.
Here's the Java implementation:
```java
class Solution {
public int singleNumber(int[] nums) {
int singleNumber = 0; // Initialize variable to store result
// Perform bitwise XOR operation on all elements in the array
for (int num : nums) {
singleNumber ^= num;
}
return singleNumber; // Return the single number
}
}
```
This implementation follows the steps outlined above and efficiently finds the single number that appears only once in the given array using bitwise XOR operation in Java.
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