[LeedCode] Single Number II

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

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

Subscribe to see which companies asked this question.

  /**   * 时间复杂度O(n)   * key:位运算   */   public class Solution {    public int singleNumber(int[] nums) {        int ones = 0, twos = 0;        for (int i = 0 ;i < nums.length;i++)        {            ones = (ones ^ nums[i]) & (~twos);            twos = (twos ^ nums[i]) & (~ones);        }        return ones;    }}

解析：
[5,5,5,1]，举例假设nums[i] = 5，二进制为nums[i] = 101;
1. 第一次碰到nums[i]，ones = nums[i] ; twos = 0 ;
ones = 0, twos = 0;
ones = (ones ^ nums[i]) & (~twos) = ( 000 ^ 101 ) & 111 = 101;
twos = (twos ^ nums[i]) & (~ones) = ( 000 ^ 101 ) & 010 = 0;
2. 第二次碰到nums[i]，ones = 0 ; twos = nums[i] ;
ones = 0, twos = nums[i];
ones = (ones ^ nums[i]) & (~twos) = ( 101 ^ 101 ) & 111 = 0;
twos = (twos ^ nums[i]) & (~ones) = ( 000 ^ 101 ) & 111 = 101;
3. 第三次碰到nums[i]，ones = 0 ; twos = 0 ;
ones = (ones ^ nums[i]) & (~twos) = ( 000 ^ 101 ) & 010 = 0;
twos = (twos ^ nums[i]) & (~ones) = ( 101 ^ 101) & 111 = 0;