[LeetCode OJ] Single Number之二 ——Given an array of integers, every element appears THREE times except for one. Find that single one.
class Solution {
public:
int singleNumber(int A[], int n) {
int bits = sizeof(int)*;
int result=;
for(int i=; i<=bits; i++)
{
int w=;
int t=; for(int j=; j<n; j++)
w += (A[j]>>(i-))&t;
result+= (w%)<<(i-); //若是除过一个数之外,其他数重复k次,则将此处的3改为k
}
return result;
}
};
利用二进制的位运算来解决这个问题,比如输入的数组是[2,2,3,2]
它们对应的二进制为:
0010 (2)
0010 (2)
0011 (3)
0010 (2)
————
对应位求和后:0 0 4 1
将所有的整数对应的二进制位加起来,再求除以3之后的余数,得到0011 (3),这就是我们要找的single number。
对于除一个整数只出现一次,其他整数都出现k次(k=2,3,4...),要求找到single number这种问题,也可以用上面位运算的方法来解决,只需将程序中的3改为k即可。
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