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?
public class SingleNumberII { public int singleNumber2(int[] nums) { //整型占4个字节，将一个整数看成32位表示的二进制数，数组arr记录了对应位上为1的个数 int[] arr = new int[32]; for(int i=0;i<nums.length;i++) { for(int j=0;j<32;j++) if((nums[i]>>j & 1)==1) { arr[j]++; } } int res=0; //数组位置0对应的数的最低位，故每次都要向左移 for(int i=0;i<arr.length;i++) { res += (arr[i]%3 <<i); } return res; } //上面是对每一个数计算一个count 这个是按照每一位对所有数计算count public int singleNumber(int[] A) { int [] count = new int[32]; int result = 0; for (int i = 0; i < 32; i++) { for (int j = 0; j < A.length; j++) { if (((A[j] >> i) & 1)==1) { count[i]++; } } result |= ((count[i] % 3) << i); } return result; } }