【Leetcode】Single Number

Given an array of integers, every element appears twice except for one. Find that single one. We can use XOR operation. Because every number XOR itself, the results will be zero. So We XOR every integer in the array, and the result is the single one we want
to find. Here is the java version code:

public class Solution {
public int singleNumber(int[] A) {
int res=0;
for(int i=0;i<A.length;i++){
res=res^A[i];
}
return res;
}
}

Follow up 1: 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? For this problem, we can't use
the XOR operation.The best way to solve this problem is use "bit count". Create a 32 length int array count[32]. count[i] means how many '1' in the ith bit of all the integers. If count[i] could be divided by 3, then we ignore this bit, else we take out this
bit and form the result.Below is java version code:

public class Solution {
public int singleNumber(int[] A) {
int res=0;
int[] count=new int[32];
for(int i=0;i<32;i++){
for(int j=0;j<A.length;j++){
if(((A[j]>>i)&1)==1){
count[i]=count[i]+1;
}
}
if((count[i]%3)!=0){
res=res|(1<<i);
}
}
return res;
}
}

Follow up 2: Given an array of integers, every element appears twice except for two. Find that two integers. Solution: First, XOR all the integers in the array we can get a result.(suppose it's c) Second, from the least significant bit to the most significant
bit, find the first '1' position(suppose the position is p). Third, divided the integers in to two groups, the p position is '1' in one group, '0' in other group. Fourth, XOR all the integers in the two groups, and the results is the two integers we want.