剑指offer（37）-数组中只出现一次的数字

题目描述

代码

class Solution {
public:
void FindNumsAppearOnce(vector<int> data,int* num1,int *num2) {
if (data.size() < 2) {
return;
}
int resultExclusiveOR = 0;
for (int i = 0; i < data.size(); i++) {
resultExclusiveOR ^= data[i];
}
unsigned int index0f1 = FindFirstBitIs1(resultExclusiveOR);
*num1 = *num2 = 0;
for (int j = 0; j < data.size(); j++) {
if (isBit1(data[j], index0f1)) {
*num1 ^= data[j];
}else{
*num2 ^= data[j];
}
}

}

// 在整数中num的二进制表示中找到最右边是1的位
unsigned int FindFirstBitIs1(int num){
int indexBit = 0;
while (((num & 1) == 0) && (indexBit < 8 * sizeof(int))) {
num = num >> 1;
++ indexBit;
}
return indexBit;
}

// 判断在num的二进制表示中从右边数起的indexBit位是不是1.
bool isBit1(int num,unsigned int indexBit){
num = num >> indexBit;
return num & 1;
}

};