/**
* Source : https://oj.leetcode.com/problems/single-number-ii/
*
* 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?
*
*/
public class SingleNumber2 {
/**
* 数组中每个元素出现3次,只有一个元素出现一次,找出出现一次的元素
*
* 沿用SingleNumber的方法,比如:{1,1,1,2,2,2,3}
* 01
* 01
* 01
* 10
* 10
* 10
* 11
* ______
* 44 对每一位求和
* 11 每一位对3求余
*
* @param arr
* @return
*/
public int singleNumber (int[] arr) {
int res = 0;
for (int i = 0; i < 32; i++) {
int sum = 0;
int mask = 1<<i;
for (int j = 0; j < arr.length; j++) {
if ((arr[j] & mask) != 0) {
sum ++;
}
}
res |= (sum % 3) << i;
}
return res;
}
public static void main(String[] args) {
SingleNumber2 singleNumber2 = new SingleNumber2();
System.out.println(singleNumber2.singleNumber(new int[]{1,1,1,2,2,2,3}) + " == 3");
System.out.println(singleNumber2.singleNumber(new int[]{14,14,14,9}) + " == 9");
}
}