标签：failed fail off cli cst 剑指offer hide ast style
// 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 <cstdio> #include <cstring> #include <cstdlib> // ====================方法一==================== // 笨方法, 时间复杂度O(nlogn) 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 > 0) { 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) { if (n <= 0) return 0; char strN[50]; sprintf_s(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‘; //输入数字首位, 字符串转数字 unsigned int length = static_cast<unsigned int>(strlen(strN)); //边界值 if (length == 1 && first == 0) //输入为 0 return 0; if (length == 1 && first > 0) //输入为 0~9 return 1; // 假设strN是 21345 //数字10000~19999中首位为1的数目 int numFirstDigit = 0; if (first > 1) numFirstDigit = PowerBase10(length - 1); //首位数字为1, 如11345, 此时10000~11345 else if (first == 1) numFirstDigit = atoi(strN + 1) + 1; //则1345 + 1为首位为1的数目. atoi转换字符串为数字 // numOtherDigits是01346-21345除了第一位之外的数位中1的数目 //此时(length - 1)表示四个位置, (length - 2)表示其余三位可能的取值0~9 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) //求10^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"); else printf("Solution1 failed.\n"); if (NumberOf1Between1AndN_Solution2(n) == expected) printf("Solution2 passed.\n"); else printf("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(); return 0; }
分析：第二种方法需要仔细分析规律。
class Solution { public: int NumberOf1Between1AndN_Solution(int n) { if (n <= 0) return 0; int number = 0; for (int i = 1; i <= n; ++i) { number += NumberOf1(i); } return number; } int NumberOf1(int n) { int number = 0; while(n) { if (n % 10 == 1) ++number; n = n / 10; } return number; } };
class Solution { public: int NumberOf1Between1AndN_Solution(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<unsigned int>(strlen(strN)); if (length == 1 && first == 0) return 0; if (length == 1 && first > 0) return 1; int numFirstDigit = 0; if (first > 1) numFirstDigit = PowerBase10(length - 1); else if (first == 1) numFirstDigit = atoi(strN + 1) + 1; int numOtherDigir = first * (length - 1) * PowerBase10(length - 2); int numRecursive = NumberOf1(strN + 1); return numFirstDigit + numOtherDigir + numRecursive; } int PowerBase10(unsigned int n) { int result = 1; for (unsigned int i = 0; i < n; ++i) result *= 10; return result; } };
标签：failed fail off cli cst 剑指offer hide ast style
原文地址：https://www.cnblogs.com/ZSY-blog/p/12632996.html
