Question

136. Single Number

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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 $O(1)$.

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

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

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

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

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

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

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

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

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

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

Solution 2

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