// 面试题43：从1到n整数中1出现的次数
// 题目：输入一个整数n，求从1到n这n个整数的十进制表示中1出现的次数。例如
// 输入12，从1到12这些整数中包含1 的数字有1，10，11和12，1一共出现了5次。

#include <iostream>
#include <cstring>
#include <cstdlib>// ====================方法一====================
//逐个判断，时间复杂度为O（nlogn），不好
int NumberOf1(unsigned int n);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;
}// ====================方法二====================
int NumberOf1(const char* strN);
int PowerBase10(unsigned int n);int NumberOf1Between1AndN_Solution2(int n)//把数字换成字符串，方便处理
{if (n <= 0)return 0;char strN[50];sprintf(strN, "%d", n);//格式化输出成字符串return NumberOf1(strN);
}int NumberOf1(const char* strN)
{if (!strN || *strN < '0' || *strN > '9' || *strN == '\0')return 0;int first = *strN - '0';//第一位的最大值unsigned int length = static_cast<unsigned int>(strlen(strN));//强制转换符if (length == 1 && first == 0)//边界特殊情况return 0;if (length == 1 && first > 0)return 1;// 假设strN是"21345"//先计算第一种情况，第一位为1的个数// 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;//若在1xx的情况，个数不到PowerBase10(length - 1)，atoi是字符串转整数//第二种情况，非第一位为1的个数// 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);//第一位可能性有first个，第二项表示选取除了第一位的任一位为1，剩下的有10种可能// numRecursive是1-1345中1的数目，使用迭代处理int numRecursive = NumberOf1(strN + 1);return numFirstDigit + numOtherDigits + numRecursive;
}int PowerBase10(unsigned int n)//10的n次方
{int result = 1;for (unsigned int i = 0; i < n; ++i)result *= 10;return result;
}// ====================测试代码====================
void Test(const char* testName, int n, int expected)
{if (testName != nullptr)printf("%s begins: \n", testName);if (NumberOf1Between1AndN_Solution1(n) == expected)printf("Solution1 passed.\n");elseprintf("Solution1 failed.\n");if (NumberOf1Between1AndN_Solution2(n) == expected)printf("Solution2 passed.\n");elseprintf("Solution2 failed.\n");printf("\n");
}void Test()
{Test("Test1", 1, 1);Test("Test2", 5, 1);Test("Test3", 10, 2);Test("Test4", 55, 16);Test("Test5", 99, 20);Test("Test6", 10000, 4001);Test("Test7", 21345, 18821);Test("Test8", 0, 0);
}int main(int argc, char* argv[])
{Test();system("pause");return 0;
}