異或操作

1至1000裏面只有一個重複，現在需要找出這個數

一個數組中只有一個數字發生了一次，其他的都是兩次，找出這個數字

一個數組中有兩個數發生了一次，其他的都是兩次，找出這兩個數

這裏這個題目都是用抑或操作來處理數組裏面的數據。

主要是下面這個公式

// NOTICE : B = A ^ (B ^ A)

第一題

直接上代碼了

#include <iostream>
#include <cstring>
#include <cstdlib>

const int N = 10;

int main()
{
    int arr[N];
    int len = sizeof(arr) / sizeof(int);
    int i = 0;
    int res = 0;
    for (i = 0; i < N; i++)
    {
        arr[i] = i + 1;
        if (i == 0)
            res = i + 1;
        else
            res ^= i + 1;
    }

    int arr_data[N+1] = {1, 3, 5, 7, 9, 10, 2, 3, 4, 6, 8};
    for (i = 0; i < N+1; i++)
        res ^= arr_data[i];
    std::cout << "only repeat number: " << res << std::endl;

    return 0;
}

上面將N改爲1000即可。

第二題

對於第二題，較爲容易，對整個數據取異或，剩下的那個值就是發生一次的數。

#include <iostream>
#include <cstring>
#include <cstdlib>

const int N = 10;

void quicksort(int arr[], int low, int high)
{
    if (low >= high)
        return;
    int i = low;
    int j = high;
    int val = arr[j];
    while (i < j)
    {
        while (i < j && arr[i] >= val)
            i++;
        arr[j] = arr[i];
        while (i < j && arr[j] < val)
            j--;
        arr[i] = arr[j];
    }
    arr[i] = val;
    quicksort(arr, low, i-1);
    quicksort(arr, i+1, high);
}

int main()
{
    int arr[] = {1, 3, 4, 3, 5, 7, 5, 9, 6, 1, 7, 9, 4};
    int len = sizeof(arr) / sizeof(int);
    int i = 0;
    std::cout << "org data:" << std::endl;
    for (i = 0; i < len; i++)
        std::cout << arr[i] << " ";
    std::cout << std::endl;

    /*
    quicksort(arr, 0, len-1);
    std::cout << "sorted data:" << std::endl;
    for (i = 0; i < len; i++)
        std::cout << arr[i] << " ";
    std::cout << std::endl;
    */

    int res = 0;
    for (i = 0; i < len; i++)
    {
        if (i == 0)
            res = arr[i];
        else
            res ^= arr[i];
    }
    std::cout << "the only number: " << res << std::endl;
    return 0;
}

第三題

對於第三題，也是一樣的，先將這個數組分組，分組以後再來獲取相應的值就可以了。

#include <iostream>
#include <cstring>
#include <cstdlib>

const int N = 10;

void quicksort(int arr[], int low, int high)
{
    if (low >= high)
        return;
    int i = low;
    int j = high;
    int val = arr[j];
    while (i < j)
    {
        while (i < j && arr[i] >= val)
            i++;
        arr[j] = arr[i];
        while (i < j && arr[j] < val)
            j--;
        arr[i] = arr[j];
    }
    arr[i] = val;
    quicksort(arr, low, i-1);
    quicksort(arr, i+1, high);
}

int main()
{
    int arr[] = {1, 3, 4, 3, 5, 7, 5, 9, 6, 1, 12, 7, 9, 4};
    int len = sizeof(arr) / sizeof(int);
    int i = 0;
    std::cout << "org data:" << std::endl;
    for (i = 0; i < len; i++)
        std::cout << arr[i] << " ";
    std::cout << std::endl;

    /*
    quicksort(arr, 0, len-1);
    std::cout << "sorted data:" << std::endl;
    for (i = 0; i < len; i++)
        std::cout << arr[i] << " ";
    std::cout << std::endl;
    */

    int res = 0;
    for (i = 0; i < len; i++)
    {
        if (i == 0)
            res = arr[i];
        else
            res ^= arr[i];
    }
    std::cout << "the split number: " << res << std::endl;
    int key = 1;
    while (1)
    {
        if (key & res)
            break;
        else
            key <<= 1;
    }
    std::cout << "the key is " << key << std::endl;
    int number1 = 0;
    int number2 = 0;
    for (i = 0; i < len; i++)
    {
        if (arr[i] & key) {
            if (number1 == 0)
                number1 = arr[i];
            else 
                number1 ^= arr[i];
        } else {
            if (number2 == 0)
                number2 = arr[i];
            else
                number2 ^= arr[i];
        }
    }
    std::cout << "the number which happens once: " << number1 << ", " << number2 << std::endl;

    return 0;
}