《劍指offer》第四十三題（從1到n整數中1出現的次數）

// 面試題43：從1到n整數中1出現的次數
// 題目：輸入一個整數n，求從1到n這n個整數的十進制表示中1出現的次數。例如
// 輸入12，從1到12這些整數中包含1 的數字有1，10，11和12，1一共出現了5次。

#include <iostream>
#include <cstring>
#include <cstdlib>// ====================方法一====================
//逐個判斷，時間復雜度為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)//把數字換成字符串，方便處理
{if (n <= 0)return 0;char strN[50];sprintf(strN, "%d", n);//格式化輸出成字符串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));//強制轉換符if (length == 1 && first == 0)//邊界特殊情況return 0;if (length == 1 && first > 0)return 1;// 假設strN是"21345"//先計算第一種情況，第一位為1的個數// numFirstDigit是數字10000-19999的第一個位中1的數目int numFirstDigit = 0;if (first > 1)numFirstDigit = PowerBase10(length - 1);else if (first == 1)numFirstDigit = atoi(strN + 1) + 1;//若在1xx的情況，個數不到PowerBase10(length - 1)，atoi是字符串轉整數//第二種情況，非第一位為1的個數// numOtherDigits是01346-21345除了第一位之外的數位中1的數目int numOtherDigits = first * (length - 1) * PowerBase10(length - 2);//第一位可能性有first個，第二項表示選取除了第一位的任一位為1，剩下的有10種可能// numRecursive是1-1345中1的數目，使用迭代處理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;
}// ====================測試代碼====================
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;
}

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