README.md

comments	difficulty	edit_url
true	中等	https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9856%20-%20I.%20%E6%95%B0%E7%BB%84%E4%B8%AD%E6%95%B0%E5%AD%97%E5%87%BA%E7%8E%B0%E7%9A%84%E6%AC%A1%E6%95%B0/README.md

面试题 56 - I. 数组中数字出现的次数

题目描述

一个整型数组 nums 里除两个数字之外，其他数字都出现了两次。请写程序找出这两个只出现一次的数字。要求时间复杂度是O(n)，空间复杂度是O(1)。

 

示例 1：

输入：nums = [4,1,4,6]
输出：[1,6] 或 [6,1]

示例 2：

输入：nums = [1,2,10,4,1,4,3,3]
输出：[2,10] 或 [10,2]

 

限制：

• 2 <= nums.length <= 10000

 

解法

方法一：位运算

由于数组中除了两个数字之外，其他数字都出现了两次，因此对数组中的所有数字进行异或运算，得到的结果即为两个只出现一次的数字的异或结果。

由于这两个数字不相等，因此异或结果中至少存在一位为 $1$。我们通过 lowbit 运算找到异或结果中最低位的 $1$，并将数组中的所有数字按照该位是否为 $1$ 分为两组，这样两个只出现一次的数字就被分到了不同的组中。

对两个组分别进行异或运算，即可得到两个只出现一次的数字。

时间复杂度 $O(n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。

Python3

class Solution:
    def singleNumbers(self, nums: List[int]) -> List[int]:
        xs = reduce(xor, nums)
        a = 0
        lb = xs & -xs
        for x in nums:
            if x & lb:
                a ^= x
        b = xs ^ a
        return [a, b]

Java

class Solution {
    public int[] singleNumbers(int[] nums) {
        int xs = 0;
        for (int x : nums) {
            xs ^= x;
        }
        int lb = xs & -xs;
        int a = 0;
        for (int x : nums) {
            if ((x & lb) != 0) {
                a ^= x;
            }
        }
        int b = xs ^ a;
        return new int[] {a, b};
    }
}

C++

class Solution {
public:
    vector<int> singleNumbers(vector<int>& nums) {
        int xs = 0;
        for (int& x : nums) {
            xs ^= x;
        }
        int lb = xs & -xs;
        int a = 0;
        for (int& x : nums) {
            if (x & lb) {
                a ^= x;
            }
        }
        int b = xs ^ a;
        return {a, b};
    }
};

Go

func singleNumbers(nums []int) []int {
	xs := 0
	for _, x := range nums {
		xs ^= x
	}
	lb := xs & -xs
	a := 0
	for _, x := range nums {
		if x&lb != 0 {
			a ^= x
		}
	}
	b := xs ^ a
	return []int{a, b}
}

TypeScript

function singleNumbers(nums: number[]): number[] {
    let xs = 0;
    for (const x of nums) {
        xs ^= x;
    }
    const lb = xs & -xs;
    let a = 0;
    for (const x of nums) {
        if (x & lb) {
            a ^= x;
        }
    }
    const b = xs ^ a;
    return [a, b];
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var singleNumbers = function (nums) {
    let xs = 0;
    for (const x of nums) {
        xs ^= x;
    }
    const lb = xs & -xs;
    let a = 0;
    for (const x of nums) {
        if (x & lb) {
            a ^= x;
        }
    }
    const b = xs ^ a;
    return [a, b];
};

C#

public class Solution {
    public int[] SingleNumbers(int[] nums) {
        int xs = 0;
        foreach(int x in nums) {
            xs ^= x;
        }
        int lb = xs & - xs;
        int a = 0;
        foreach(int x in nums) {
            if ((x & lb) != 0) {
                a ^= x;
            }
        }
        int b = xs ^ a;
        return new int[] {a, b};
    }
}

Swift

class Solution {
    func singleNumbers(_ nums: [Int]) -> [Int] {
        var xorSum = 0
        for num in nums {
            xorSum ^= num
        }

        let lowBit = xorSum & -xorSum
        var a = 0
        for num in nums {
            if (num & lowBit) != 0 {
                a ^= num
            }
        }

        let b = xorSum ^ a
        return [a, b]
    }
}