只出现一次的数字 II

137. 只出现一次的数字 II

给你一个整数数组 nums ，除某个元素仅出现一次外，其余每个元素都恰出现 三次。请你找出并返回那个只出现了一次的元素。

示例 1：

输入：nums = [2,2,3,2]
输出：3

示例 2：

输入：nums = [0,1,0,1,0,1,99]
输出：99

提示：

1 <= nums.length <= 3 * 10⁴
-2³¹ <= nums[i] <= 2³¹ - 1
nums 中，除某个元素仅出现一次外，其余每个元素都恰出现三次

进阶：你的算法应该具有线性时间复杂度。你可以不使用额外空间来实现吗？

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class Solution { public: int singleNumber(vector<int>& nums) { } };

rust 解法, 执行用时: 0 ms, 内存消耗: 2.1 MB, 提交时间: 2023-10-15 11:06:56

impl Solution {
    pub fn single_number1(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        for i in 0..32 {
            let mut cnt1 = 0;
            for &x in &nums {
                cnt1 += x >> i & 1;
            }
            ans |= cnt1 % 3 << i;
        }
        ans
    }

    pub fn single_number2(nums: Vec<i32>) -> i32 {
        let mut a = 0;
        let mut b = 0;
        for &x in &nums {
            let tmp_a = a;
            a = (a ^ x) & (a | b);
            b = (b ^ x) & !tmp_a;
        }
        b
    }

    pub fn single_number(nums: Vec<i32>) -> i32 {
        let mut a = 0;
        let mut b = 0;
        for &x in &nums {
            b = (b ^ x) & !a;
            a = (a ^ x) & !b;
        }
        b
    }
}

javascript 解法, 执行用时: 60 ms, 内存消耗: 42.8 MB, 提交时间: 2023-10-15 11:06:12

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


var singleNumber2 = function(nums) {
    let a = 0, b = 0;
    for (const x of nums) {
        [a, b] = [(a ^ x) & (a | b), (b ^ x) & ~a]
    }
    return b;
};

var singleNumber3 = function(nums) {
    let ans = 0;
    for (let i = 0; i < 32; i++) {
        let cnt1 = 0;
        for (const x of nums) {
            cnt1 += x >> i & 1;
        }
        ans |= cnt1 % 3 << i;
    }
    return ans;
};

golang 解法, 执行用时: 4 ms, 内存消耗: 3.3 MB, 提交时间: 2023-10-15 11:05:32

func singleNumber1(nums []int) int {
    ans := int32(0)
    for i := 0; i < 32; i++ {
        cnt1 := int32(0)
        for _, x := range nums {
            cnt1 += int32(x) >> i & 1
        }
        ans |= cnt1 % 3 << i
    }
    return int(ans)
}

func singleNumber2(nums []int) int {
    a, b := 0, 0
    for _, x := range nums {
        a, b = (a^x)&(a|b), (b^x)&^a
    }
    return b
}

func singleNumber(nums []int) int {
    a, b := 0, 0
    for _, x := range nums {
        b = (b ^ x) &^ a
        a = (a ^ x) &^ b
    }
    return b
}

cpp 解法, 执行用时: 4 ms, 内存消耗: 9.5 MB, 提交时间: 2023-10-15 11:04:48

class Solution {
public:
    int singleNumber1(vector<int> &nums) {
        int ans = 0;
        for (int i = 0; i < 32; i++) {
            int cnt1 = 0;
            for (int x: nums) {
                cnt1 += x >> i & 1;
            }
            ans |= cnt1 % 3 << i;
        }
        return ans;
    }

    int singleNumber2(vector<int> &nums) {
        int a = 0, b = 0;
        for (int x: nums) {
            int tmp_a = a;
            a = (a ^ x) & (a | b);
            b = (b ^ x) & ~tmp_a;
        }
        return b;
    }

    int singleNumber(vector<int> &nums) {
        int a = 0, b = 0;
        for (int x: nums) {
            b = (b ^ x) & ~a;
            a = (a ^ x) & ~b;
        }
        return b;
    }
};

java 解法, 执行用时: 0 ms, 内存消耗: 42.9 MB, 提交时间: 2023-10-15 11:03:59

class Solution {
    public int singleNumber1(int[] nums) {
        int ans = 0;
        for (int i = 0; i < 32; i++) {
            int cnt1 = 0;
            for (int x : nums) {
                cnt1 += x >> i & 1;
            }
            ans |= cnt1 % 3 << i;
        }
        return ans;
    }

    public int singleNumber2(int[] nums) {
        int a = 0, b = 0;
        for (int x : nums) {
            int tmpA = a;
            a = (a ^ x) & (a | b);
            b = (b ^ x) & ~tmpA;
        }
        return b;
    }

    public int singleNumber(int[] nums) {
        int a = 0, b = 0;
        for (int x : nums) {
            b = (b ^ x) & ~a;
            a = (a ^ x) & ~b;
        }
        return b;
    }
}

python3 解法, 执行用时: 48 ms, 内存消耗: 14.4 MB, 提交时间: 2020-11-19 01:51:50

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

python3 解法, 执行用时: 36 ms, 内存消耗: 14.6 MB, 提交时间: 2020-11-18 22:16:49

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        counter = collections.Counter(nums)
        for (i, k) in counter.items():
            if k == 1:
                return i

python3 解法, 执行用时: 48 ms, 内存消耗: 14.8 MB, 提交时间: 2020-11-18 22:14:45

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        return  (3 * sum(set(nums)) - sum(nums)) // 2

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