Single number II

Single number II Total accepted: 20813 total submissions: 62103my submissions

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

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

Discuss

Because all other numbers appear three times, the sum of 1 on the nth bit of all numbers must be a multiple of 3, or % 3 is 1. Because if there is no single number, the sum of 1 on the nth bit of all numbers must be a multiple of 3. After a single number is added, if the number is 1 on this, the sum of % 3 is 1. Otherwise, if the value is zero, it is a multiple of 3.

It applies to all other numbers that appear n> = 3 times, and only one number appears once.

public class Solution {    public int singleNumber(int[] A) {       int arr[] = new int[33];        //and the number on every bit,and add the bits       for(int i=0;i<=31;i++)       {           int bit = 1<<i;           // the bits is 2^31           for(int j=0;j<A.length;j++)           {               // this must be unequal, not bigger than               if((A[j]&bit)!=0)               {                   arr[i]++;               }           }       }       int res = 0;       // reconstruct the number       for(int i=0;i<=31;i++)       {           if(arr[i]%3!=0)           {               res += 1<<i;           }       }        return res;    }}