Given an array of integers, every element appears three times except for one. Find that single one.
public class Solution { public int singleNumber(int[] nums) { //int的32个bit逐个处理,每一个bit进行相加,然后mod3,即可判断多余的一位是1或0.此思想同样适用于均出现三次,一个出现两次场景 int res=0; for(int i=0;i<32;i++){ int count=0; for(int j=0;j<nums.length;j++){ count+=(nums[j]>>i)&1; } // res+=(count%3)<<i;//均可 res|=(count%3)<<i; } return res; } //http://blog.csdn.net/kenden23/article/details/13625297 }