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.

 

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:
        bits = [0] * 32
        for num in nums:
            for i in range(32):
                bits[i] += (num & 1)
                num >>= 1
        res = 0
        for i in range(32):
            if bits[i] % 3 != 0:
                res |= (1 << i)
        return res if bits[31] % 3 == 0 else ~(res ^ 0xffffffff)

Java

class Solution {
    public int singleNumber(int[] nums) {
        int[] bits = new int[32];
        for (int num : nums) {
            for (int i = 0; i < 32; ++i) {
                bits[i] += (num & 1);
                num >>= 1;
            }
        }

        int res = 0;
        for (int i = 0; i < 32; ++i) {
            if (bits[i] % 3 == 1) {
                res |= (1 << i);
            }
        }
        return res;
    }
}

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