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NumberOf1.cpp
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141 lines (109 loc) · 3.18 KB
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/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 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====================
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)
{
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;
// 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;
}
// ====================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;
}