JZ-C-32

剑指offer第三十二题：从1到n整数中1出现的次数

//============================================================================
// Name        : JZ-C-32.cpp
// Author      : Laughing_Lz
// Version     :
// Copyright   : All Right Reserved
// Description : 从1到n整数中1出现的次数
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;

// ====================方法一:未考虑时间效率，时间复杂度O(n*logn)====================
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); //对每个数字都要做除法和求余运算

    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：将数字分为两段，如21345分为从1到1345 和 1346到21345，时间复杂度O(logn)====================
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); //把格式化的数据写入某个字符串，这里是将整数写入strN字符数组中

    return NumberOf1(strN);
}

int NumberOf1(const char* strN) {
    if (!strN || *strN < '0' || *strN > '9' || *strN == '')
        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;    //atoi函数：将字符串转成整数

    // 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);    //★★
    // 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(char* testName, int n, int expected) {
    if (testName != NULL)
        printf("%s begins: 
", testName);

    if (NumberOf1Between1AndN_Solution1(n) == expected)
        printf("Solution1 passed.
");
    else
        printf("Solution1 failed.
");

    if (NumberOf1Between1AndN_Solution2(n) == expected)
        printf("Solution2 passed.
");
    else
        printf("Solution2 failed.
");

    printf("
");
}

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;
}