在以二进制格式表示的整数数组中查找缺少的元素

给定N个字符串，该字符串以二进制格式表示从0到N的所有整数，除了任何一个。任务是找到丢失的号码。输入由字符串数组组成，其中数组元素以二进制格式表示。

例子：

Input: arr[] = {“0000”, “0001”, “0010”, “0100”}
Output: 3

Input: arr[] = {“0000”, “0001”, “0010”, “0011”, “0100”, “0110”, “0111”, “1000”}
Output: 5

方法：

• 在给定的N个整数中，可以看到数字的最低有效位的1和0的不平衡。由于缺少一个数字，因此LSB中的0或1都不存在。如果缺少的数字的LSB = 0，则count(1)将大于等于count(0)。如果缺少数字的LSB为1，则count(1)小于count(0)。
• 从步骤1可以轻松确定遗漏号码的LSB。
• 确定后，丢弃所有LSB与丢失数字不同的数字，即，如果丢失数字的LSB = 0，则丢弃所有LSB = 1的数字，反之亦然。
• 再次从第1步继续执行该过程，然后重复进行下一个LSB。
• 继续上述过程，直到遍历所有位。

以下是上述方法的实现：

// C++ program to find the missing integer
// in N numbers when N bits are given
#include 
using namespace std;
  
class BitInteger {
private:
    bool* bits;
  
public:
    static const int INTEGER_SIZE = 32;
  
    BitInteger()
    {
        bits = new bool[INTEGER_SIZE];
    }
  
    // Constructor to convert an integer
    // variable into binary format
    BitInteger(int value)
    {
        bits = new bool[INTEGER_SIZE];
  
        for (int j = 0; j < INTEGER_SIZE; j++) {
  
            // The if statement will shift the
            // original value j times.
            // So that appropriate (INTEGER_SIZE - 1 -j)th
            // bits will be either 0/1.
            //  (INTEGER_SIZE - 1 -j)th bit for all
            // j = 0 to INTEGER_SIZE-1 corresponds
            // to  LSB to MSB respectively.
            if (((value >> j) & 1) == 1)
                bits[INTEGER_SIZE - 1 - j] = true;
            else
                bits[INTEGER_SIZE - 1 - j] = false;
        }
    }
    // Constructor to convert a
    // string into binary format.
    BitInteger(string str)
    {
        int len = str.length();
        int x = INTEGER_SIZE - len;
        bits = new bool[INTEGER_SIZE];
  
        // If len = 4. Then x = 32 - 4 = 28.
        // Hence iterate from
        // bit 28 to bit 32 and just
        // replicate the input string.
        int i = 0;
  
        for (int j = x; j <= INTEGER_SIZE && i < len; j++, i++) {
            if (str[i] == '1')
                bits[j] = true;
            else
                bits[j] = false;
        }
    }
  
    // this function fetches the kth bit
    int fetch(int k)
    {
        if (bits[k])
            return 1;
  
        return 0;
    }
  
    // this function will set a value
    // of bit indicated by k to given bitValue
    void set(int k, int bitValue)
    {
        if (bitValue == 0)
            bits[k] = false;
        else
            bits[k] = true;
    }
  
    // convert binary representation to integer
    int toInt()
    {
        int n = 0;
        for (int i = 0; i < INTEGER_SIZE; i++) {
            n = n << 1;
            if (bits[i])
                n = n | 1;
        }
        return n;
    }
};
  
// Function to find the missing number
int findMissingFunc(list& myList, int column)
{
    // This means that we have processed
    // the entire 32 bit binary number.
    if (column < 0)
        return 0;
  
    list oddIndices;
    list evenIndices;
  
    for (BitInteger t : myList) {
  
        // Initially column = LSB. So
        // if LSB of the given number is 0,
        // then the number is even and
        // hence we add it to evenIndices list.
        // else if LSB = 0 then add it to oddIndices list.
        if (t.fetch(column) == 0)
            evenIndices.push_back(t);
        else
            oddIndices.push_back(t);
    }
  
    // Step 1 and Step 2 of the algorithm.
    // Here we determine the LSB bit of missing number.
  
    if (oddIndices.size() >= evenIndices.size())
  
        // LSB of the missing number is 0.
        // Hence it is an even number.
        // Step 3 and 4 of the algorithm
        // (discarding all odd numbers)
        return (findMissingFunc(evenIndices, column - 1)) << 1 | 0;
  
    else
        // LSB of the missing number is 1.
        // Hence it is an odd number.
        // Step 3 and 4 of the algorithm
        // (discarding all even numbers)
        return (findMissingFunc(oddIndices, column - 1)) << 1 | 1;
}
  
// Function to return the missing integer
int findMissing(list& myList)
{
    // Initial call is with given array and LSB.
    return findMissingFunc(myList, BitInteger::INTEGER_SIZE - 1);
}
  
// Driver Code.
int main()
{
  
    // a corresponds to the input array which
    // is a list of binary numbers
    list a = { BitInteger("0000"), BitInteger("0001"),
                           BitInteger("0010"), BitInteger("0100"),
                           BitInteger("0101") };
    int missing1 = findMissing(a);
    cout << missing1 << "\n";
  
    return 0;
}

输出：

3

时间复杂度： O(N)