README_EN.md

137. Single Number II

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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.

 

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.

 

Follow up: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

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

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;
    }
}

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)
}

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;
    }
};

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