数组中只出现一次的数字

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问题:一个整型数组里除了两个数字之外，其他的数字都出现了两次。请写程序找出这两个只出现一次的数字。要求时间复杂度是O(n),空间复杂度为O(1).

算法思想:采取异或操作和根据结果二进制中1的位置进行划分为两个部分，然后在采取异或运算得出不同的结果。

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1bool 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;}

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