面试题40之数组中只出现一次的数字_NumbersAppearOnce.cpp

// NumbersAppearOnce.cpp : Defines the entry point for the console application.
//

// 《剑指Offer——名企面试官精讲典型编程题》代码
// 著作权所有者：何海涛

#include "stdafx.h"

unsigned int FindFirstBitIs1(int num);
bool IsBit1(int num, unsigned int indexBit);

void FindNumsAppearOnce(int data[], int length, int* num1, int* num2)
{
    if (data == NULL || length < 2)
        return;
 
    int resultExclusiveOR = 0;
    for (int i = 0; i < length; ++ i)
        resultExclusiveOR ^= data[i];
 
    unsigned int indexOf1 = FindFirstBitIs1(resultExclusiveOR);    
 
    *num1 = *num2 = 0;
    for (int j = 0; j < length; ++ j)
    {
        if(IsBit1(data[j], indexOf1))
            *num1 ^= data[j];
        else
            *num2 ^= data[j];
    }
}
 
// 找到num从右边数起第一个是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;
}

// 判断数字num的第indexBit位是不是1
bool IsBit1(int num, unsigned int indexBit)
{
    num = num >> indexBit;
    return (num & 1);
}

// ====================测试代码====================
void Test(char* testName, int data[], int length, int expected1, int expected2)
{
    if(testName != NULL)
        printf("%s begins: ", testName);

    int result1, result2;
    FindNumsAppearOnce(data, length, &result1, &result2);

    if((expected1 == result1 && expected2 == result2) ||
        (expected2 == result1 && expected1 == result2))
        printf("Passed.\n\n");
    else
        printf("Failed.\n\n");
}

void Test1()
{
    int data[] = {2, 4, 3, 6, 3, 2, 5, 5};
    Test("Test1", data, sizeof(data) / sizeof(int), 4, 6);
}

void Test2()
{
    int data[] = {4, 6};
    Test("Test2", data, sizeof(data) / sizeof(int), 4, 6);
}

void Test3()
{
    int data[] = {4, 6, 1, 1, 1, 1};
    Test("Test3", data, sizeof(data) / sizeof(int), 4, 6);
}

int _tmain(int argc, _TCHAR* argv[])
{
    Test1();
    Test2();
    Test3();

    return 0;
}