【LeetCode】SingleNumberII_137

package com.leetCode;

/**
* 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?
*
* @author Zealot
* @date 2015年7月25日 下午6:16:12
*/
public class SingleNumberII_137 {
public int singleNumber(int[] nums) {
int ones = 0, twos = 0;
for(int i = 0; i < nums.length; i++){
ones = (ones ^ nums[i]) & ~twos;
twos = (twos ^ nums[i]) & ~ones;
}
return ones;
}
//遍历数组中每一个元素
//计算每一个元素中，2进制中每一位加到一个总的数组里边，都加到一起之后，再与3取余，因为除了一个之外，都是出现了3次，只有那一个之出现了不到3次
//最后数组转成数字
public int singleNumber2(int[] A) {
if(A == null || A.length == 0) return 0;
int[] a = new int[32];
for(int i = 0; i < A.length; i++) {
for(int j = 0; j < 32; j++) {
if((A[i] & (1 << j)) != 0)
a[j] = (a[j] + 1) % 3;
}
}
int result = 0;
for(int i = 0; i < 32; i++) {
if(a[i] > 0)
result |= (a[i] << i);
}
return result;
}
public static void main(String[] args) {
SingleNumberII_137 s = new SingleNumberII_137();
int[] nums = {1,2,3,3,3,2,5,2,1,1};
System.out.println(s.singleNumber2(nums));
//		System.out.println(77>>1);
//		System.out.println(4>>1);
//		System.out.println(Integer.toBinaryString(4).length());
//		System.out.println("123456".length());
}
}