【剑指Offer】56.1 数组中只出现一次的数字，其他出现2次

//num1,num2分别为长度为1的数组。传出参数
//将num1[0],num2[0]设置为返回结果
public class Solution {
    public void FindNumsAppearOnce(int[] nums, int num1[], int num2[]) {
        int diff = 0;
        for (int num : nums)
            diff ^= num;
        diff &= -diff;
        for (int num : nums) {
            if ((num & diff) == 0)
                num1[0] ^= num;
            else
                num2[0] ^= num;
        }
    }
}

//num1,num2分别为长度为1的数组。传出参数
//将num1[0],num2[0]设置为返回结果
public class Solution {
    public void FindNumsAppearOnce(int[] nums, int num1[], int num2[]) {
        int diff = 0;
        for (int num : nums)
            diff ^= num;
        int index = 0;
        // 找到sum为1的位数
        while((diff >> index & 1) == 0) index++;
        
        int first = 0;
        for(int i : nums)
            if((i >> index & 1) == 0) 
                first ^= i;
        int second = diff ^= first;
        
        num1[0] = first;
        num2[0] = second;
    }
}