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?
对于数组{1,1,1,2}来说，分析每一位：
代码语言：javascript复制0 0 0 1
0 0 0 1
0 0 0 1
0 0 1 0
……………
0 0 1 3 (位数)可以分析，分别计算每一位1的出现次数n，如果n%3==1，那么结果中这一位就应该是1。
代码语言：javascript复制 public int singleNumber(int[] nums) {
int rt = 0, bit = 1;
for (int i = 0; i < 32; i ) {
int count = 0;
for (int v : nums) {
if ((v & bit) != 0) {
count ;
}
}
bit <<= 1;
rt |= (count % 3) << i;
}
return rt;
}
