10.4 You have an array with all the numbers from 1 to N, where N is at most 32,000. The array may have duplicate entries and you do not know what N is. With only 4 kilobytes of memory available, how would you print all duplicate elements in the array?

这道题给了我们很多在区间[1, 32000]中的数,让我们只用4KB的内存大小来找出所有的重复项。跟之前那道题很类似10.3 Integer not Contain in the File 文件中不包含的数,还是需要用位向量Bit Vector来解,4KB内存共有215个位Bit,大于32000个数,所以我们可以用每个bit来表示一个数,我们需要一个位向量类BitSet,来建立和数字之间的映射,有点像哈希表的功能,是一个整型数组,由于int型最大可以表示232,所以一个位置就可以映射232个数字,参见代码如下:

class BitSet {
public:
vector<int> _bitset;
BitSet(int size) {
_bitset.resize((size >> ) + );
}
bool get(int pos) {
int wordNum = (pos >> );
int bitNum = (pos % );
return (_bitset[wordNum] & ( << bitNum)) != ;
}
void set(int pos) {
int wordNum = (pos >> );
int bitNum = (pos % );
_bitset[wordNum] |= << bitNum;
}
}; class Solution {
public:
void checkDuplicates(vector<int> array) {
BitSet *bs = new BitSet();
for (int i = ; i < array.size(); ++i) {
int num = array[i];
int num0 = num - ;
if (bs->get(num0)) {
cout << num << endl;
} else {
bs->set(num0);
}
}
}
};

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