Question

Formatted question description: https://leetcode.ca/all/137.html

137	Single Number II

Given an array of integers, every element appears three times except for one. Find that single one.

Example 1:

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

Example 2:

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

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

@tag-bit

Algorithm

A 32-bit number can be created to count the number of occurrences of 1 in each digit.

If a certain digit is 1, then if the integer appears three times, the remainder of 3 is 0, so that the numbers of each digit position are added up then take the remainder of 3, and the final number remaining is a single number.

Code

Java

public class Single_Number_II {

    public class Solution {
        public int singleNumber(int[] nums) {

            if (nums == null || nums.length == 0) {
                return 0;
            }

            // first doubt, how long bits should I check.
            // int is 32 bit, so bitmap length is 32
            int single = 0;
            for (int bitPos = 0; bitPos < 32; bitPos++) {
                int sumAtThisPos = 0;

                for (int j = 0; j < nums.length; j++) {
                    int digitInA = ((nums[j] >> bitPos) & 1);
                    sumAtThisPos += digitInA;
                }

                int afterMod = sumAtThisPos % 3;
                single |= (afterMod << bitPos);
            }

            return single;
        }
    }
}

// OJ: https://leetcode.com/problems/single-number-ii/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int singleNumber(vector<int>& nums) {
        unsigned ans = 0;
        for (int i = 0; i < 32; ++i) {
            int cnt = 0;
            for (unsigned n : nums) {
                cnt = (cnt + ((n >> i) & 1)) % 3;
            }
            ans |= cnt << i;
        }
        return ans;
    }
};

class Solution(object):
  def singleNumber(self, nums):
    """
    :type nums: List[int]
    :rtype: int
    """

    def singleNumberK(nums, k):
      ret = 0
      count = [0] * 32
      for i in range(0, 32):
        for num in nums:
          if num & (1 << i):
            count[i] += 1
        if count[i] % 3 != 0:
          ret |= 1 << i
      if ret > 0x7fffffff:
        ret -= 0x100000000
      return ret

    return singleNumberK(nums, 3)

137 - Single Number II

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