136. Single Number && 137. Single Number II && 260. Single Number III

136. Single Number

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

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

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Hash Table Bit Manipulation

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(M) Single Number II (M) Single Number III (M) Missing Number (H) Find the Duplicate Number

public class Solution {
    public int singleNumber(int[] nums) {
        int xor = 0;
        for(int n : nums)
            xor ^= n;
        return xor;
    }
}

137. Single Number II

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

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

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Bit Manipulation

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(M) Single Number (M) Single Number III

...

260. Single Number III

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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Bit Manipulation

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(M) Single Number (M) Single Number II

public class Solution {
    public int[] singleNumber(int[] nums) {
        // Pass 1 :
        // Get the XOR of the two numbers we need to find
        int diff = 0;
        for (int num : nums) {
            diff ^= num;
        }
        // Get the rightmost different bit between these two numbers.
        diff &= -diff;

        // Pass 2 :
        int[] rets = {0, 0}; //split the two numbers we are looking for into two groups.
        for (int num : nums)
        {
            if ((num & diff) == 0) // the group that has rightmost bit unset.
            {
                rets[0] ^= num;
            }
            else // the group that has rightmost bit set.
            {
                rets[1] ^= num;
            }
        }
        return rets;
    }
}