README_EN.md

137. Single Number II

中文文档

Description

Given an integer array nums where every element appears three times except for one, which appears exactly once. Find the single element and return it.

You must implement a solution with a linear runtime complexity and use only constant extra space.

 

Example 1:

Input: nums = [2,2,3,2]
Output: 3

Example 2:

Input: nums = [0,1,0,1,0,1,99]
Output: 99

 

Constraints:

• 1 <= nums.length <= 3 * 104
• -231 <= nums[i] <= 231 - 1
• Each element in nums appears exactly three times except for one element which appears once.

Solutions

Python3

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        ans = 0
        for i in range(32):
            cnt = sum(num >> i & 1 for num in nums)
            if cnt % 3:
                if i == 31:
                    ans -= 1 << i
                else:
                    ans |= 1 << i
        return ans

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        a = b = 0
        for c in nums:
            aa = (~a & b & c) | (a & ~b & ~c)
            bb = ~a & (b ^ c)
            a, b = aa, bb
        return b

Java

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

class Solution {
    public int singleNumber(int[] nums) {
        int a = 0, b = 0;
        for (int c : nums) {
            int aa = (~a & b & c) | (a & ~b & ~c);
            int bb = ~a & (b ^ c);
            a = aa;
            b = bb;
        }
        return b;
    }
}

Go

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

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

C++

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

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int a = 0, b = 0;
        for (int c : nums) {
            int aa = (~a & b & c) | (a & ~b & ~c);
            int bb = ~a & (b ^ c);
            a = aa;
            b = bb;
        }
        return b;
    }
};

TypeScript

function singleNumber(nums: number[]): number {
    let ans = 0;
    for (let i = 0; i < 32; i++) {
        const count = nums.reduce((r, v) => r + ((v >> i) & 1), 0);
        ans |= count % 3 << i;
    }
    return ans;
}

function singleNumber(nums: number[]): number {
    let a = 0;
    let b = 0;
    for (const c of nums) {
        const aa = (~a & b & c) | (a & ~b & ~c);
        const bb = ~a & (b ^ c);
        a = aa;
        b = bb;
    }
    return b;
}

Rust

impl Solution {
    pub fn single_number(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        for i in 0..32 {
            let count = nums.iter().map(|v| v >> i & 1).sum::<i32>();
            ans |= count % 3 << i;
        }
        ans
    }
}

C

int singleNumber(int* nums, int numsSize) {
    int ans = 0;
    for (int i = 0; i < 32; i++) {
        int count = 0;
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] >> i & 1) {
                count++;
            }
        }
        ans |= (uint) (count % 3) << i;
    }
    return ans;
}

Swift

class Solution {
    func singleNumber(_ nums: [Int]) -> Int {
        var a = nums.sorted()
        var n = a.count
        for i in stride(from: 0, through: n - 2, by: 3) {
            if a[i] != a[i + 1] {
                return a[i]
            }
        }
        return a[n - 1]
    }
}

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