本文共 1365 字，大约阅读时间需要 4 分钟。

int NumberOf1Between1AndN_Solution1(unsigned int n){    int number = 0;    for(unsigned int i = 1; i <= n; ++ i)        number += NumberOf1(i);    return number;}int NumberOf1(unsigned int n){    int number = 0;    while(n)    {        if(n % 10 == 1)            number ++;        n = n / 10;    }    return number;}

int NumberOf1Between1AndN_Solution2(int n){    if(n <= 0)        return 0;    char strN[50];    sprintf(strN, "%d", n);    return NumberOf1(strN);}int NumberOf1(const char* strN){    if(!strN || *strN < '0' || *strN > '9' || *strN == '\0')        return 0;    int first = *strN - '0';    unsigned int length = static_cast
   
    (strlen(strN));    if(length == 1 && first == 0)        return 0;    if(length == 1 && first > 0)        return 1;    // 假设strN是"21345"    // numFirstDigit是数字10000-19999的第一个位中1的数目    int numFirstDigit = 0;    if(first > 1)        numFirstDigit = PowerBase10(length - 1);    else if(first == 1)        numFirstDigit = atoi(strN + 1) + 1;    // numOtherDigits是01346-21345除了第一位之外的数位中1的数目    int numOtherDigits = first * (length - 1) * PowerBase10(length - 2);    // numRecursive是1-1345中1的数目    int numRecursive = NumberOf1(strN + 1);    return numFirstDigit + numOtherDigits + numRecursive;}int PowerBase10(unsigned int n){    int result = 1;    for(unsigned int i = 0; i < n; ++ i)        result *= 10;    return result;}