《剑指offer》第四十五题：把数组排成最小的数

// 面试题45：把数组排成最小的数
// 题目：输入一个正整数数组，把数组里所有数字拼接起来排成一个数，打印能拼
// 接出的所有数字中最小的一个。例如输入数组{3, 32, 321}，则打印出这3个数
// 字能排成的最小数字321323。

#include "cstdio"
#include <string>
#include <algorithm>

#pragma warning(disable:4996) //忽略安全性较低的vs警告

int compare(const void* strNumber1, const void* strNumber2);

// int型整数用十进制表示最多只有10位
const int g_MaxNumberLength = 10;

char* g_StrCombine1 = new char[g_MaxNumberLength * 2 + 1];
char* g_StrCombine2 = new char[g_MaxNumberLength * 2 + 1];

void PrintMinNumber(const int* numbers, int length)
{
    if (numbers == nullptr || length <= 0)
        return;

    //将数组转换为字符串, 二维数组注意**
    char** strNumber = (char**)(new int[length]);
    for (int i = 0; i < length; ++i)
    {
        strNumber[i] = new char[g_MaxNumberLength + 1];
        sprintf(strNumber[i], "%d", numbers[i]);
    }

    qsort(strNumber, length, sizeof(char*), compare);

    for (int i = 0; i < length; ++i)
        printf("%s", strNumber[i]);
    printf("\n");

    for (int i = 0; i < length; ++i)  //记得删除创建的二维数组
        delete[] strNumber[i];
    delete[] strNumber;
}

// 如果[strNumber1][strNumber2] > [strNumber2][strNumber1], 返回值大于0
// 如果[strNumber1][strNumber2] = [strNumber2][strNumber1], 返回值等于0
// 如果[strNumber1][strNumber2] < [strNumber2][strNumber1], 返回值小于0
int compare(const void* strNumber1, const void* strNumber2)
{
    //[strNumber1][strNumber2]
    strcpy(g_StrCombine1, *(const char**)strNumber1);
    strcat(g_StrCombine1, *(const char**)strNumber2);

    //[strNumber2][strNumber1]
    strcpy(g_StrCombine2, *(const char**)strNumber2);
    strcat(g_StrCombine2, *(const char**)strNumber1);

    return strcmp(g_StrCombine1, g_StrCombine2);
}

// ====================测试代码====================
void Test(const char* testName, int* numbers, int length, const char* expectedResult)
{
    if (testName != nullptr)
        printf("%s begins:\n", testName);

    if (expectedResult != nullptr)
        printf("Expected result is: \t%s\n", expectedResult);

    printf("Actual result is: \t");
    PrintMinNumber(numbers, length);

    printf("\n");
}

void Test1()
{
    int numbers[] = { 3, 5, 1, 4, 2 };
    Test("Test1", numbers, sizeof(numbers) / sizeof(int), "12345");
}

void Test2()
{
    int numbers[] = { 3, 32, 321 };
    Test("Test2", numbers, sizeof(numbers) / sizeof(int), "321323");
}

void Test3()
{
    int numbers[] = { 3, 323, 32123 };
    Test("Test3", numbers, sizeof(numbers) / sizeof(int), "321233233");
}

void Test4()
{
    int numbers[] = { 1, 11, 111 };
    Test("Test4", numbers, sizeof(numbers) / sizeof(int), "111111");
}

// 数组中只有一个数字
void Test5()
{
    int numbers[] = { 321 };
    Test("Test5", numbers, sizeof(numbers) / sizeof(int), "321");
}

void Test6()
{
    Test("Test6", nullptr, 0, "Don't print anything.");
}


int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();
    Test6();

    return 0;
}

分析：快排算法。

class Solution {
public:
    string PrintMinNumber(vector<int> numbers) {
        
        int length = (int) numbers.size();
        string strNumber = "";
        
        sort(numbers.begin(), numbers.end(), compare);
        
        for (int i = 0; i < length; ++i)
            strNumber += to_string(numbers[i]);
        
        return strNumber;
    }
    static bool compare(int strNumber1, int strNumber2)
    {
        string g_StrCombine1 = "";
        string g_StrCombine2 = "";
        
        //[strNumber1][strNumber2]
        g_StrCombine1 += to_string(strNumber1);
        g_StrCombine1 += to_string(strNumber2);

        //[strNumber2][strNumber1]
        g_StrCombine2 += to_string(strNumber2);
        g_StrCombine2 += to_string(strNumber1);

        return g_StrCombine1 < g_StrCombine2;
    }
};