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

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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