README_EN.md

136. Single Number

中文文档

Description

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

Follow up: Could you implement a solution with a linear runtime complexity and without using extra memory?

 

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

Python3

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        res = 0
        for num in nums:
            res ^= num
        return res

Java

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

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var singleNumber = function (nums) {
    let res = 0;
    for (let num of nums) {
        res ^= num;
    }
    return res;
};

Go

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

C++

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

Rust

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

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