# LeetCode Solution, Easy, 136. Single Number

# [136. Single Number](https://leetcode.com/problems/single-number/)

### 題目敘述

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 * 10**4`
- `-3 * 10**4 <= nums[i] <= 3 * 10**4`
- Each element in the array appears twice except for one element which appears only once.

#### 題目翻譯

給定一個整數的陣列 `nums`，其中的元素都是兩兩成對的加上一個單一的數值。從陣列 `nums` 中找出是哪一個數值是只有單獨的一個。(情人節剛結束出這題是想嘲諷單身狗？！

### 解法解析

這題讓我想到前幾天的 [LeetCode Solution, Easy, 389. Find the Difference](https://blog.taiwolskit.com/leetcode-solution-easy-389-find-the-difference) 的解題邏輯一樣，找出不同的數值。所以使用 Bit operator 是最快速的方式，讓 Time complexity 可以為 `O(n)`，且 Space complexity `O(1)`。這邊唯一不同的是 PHP，因為有內建的計算函數，所以直接使用。

#### 程式範例

##### Go

```go
func singleNumber(nums []int) int {
	res := 0
	for _, n := range nums {
		res = res ^ n
	}
	return res
}
```

##### JavaScript

```javascript
/**
 * @param {number[]} nums
 * @return {number}
 */
var singleNumber = function (nums) {
    let result = 0;
    for (let i of nums) {
        result ^= i;
    }
    return result;
};
```

##### Kotlin

```kotlin
class Solution {
    fun singleNumber(nums: IntArray): Int {
        var result: Int = 0
        for (i in nums) {
            result = result xor i
        }
        return result
    }
}
```

##### PHP

```php
class Solution
{

    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function singleNumber($nums)
    {
        $count_nums = array_count_values($nums);
        foreach ($count_nums as $key => $value) {
            if ($value == 1) {
                return $key;
            }
        }
    }
}
```

##### Python

```python
class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        """
        :type nums: List[int]
        :rtype: int
        """
        a = 0
        for i in nums:
            a ^= i
        return a
```

##### Rust

```rust
impl Solution {
    pub fn single_number(nums: Vec<i32>) -> i32 {
        let mut res = 0;
        for num in nums {
            res ^= num;
        }
        res
    }
}
```

##### Swift

```swift
class Solution {
    func singleNumber(_ nums: [Int]) -> Int {
        var sum = 0
        for num in nums {
            sum ^= num
        }
        return sum
    }
}
```