查找由数组的数字形成的最大数字

如何解决查找由数组的数字形成的最大数字

任务是找到由数组的数字形成的最大数字。例如：

输入：7、56、94、1 输出：947561

在代码中，我比较ab和ba，然后使用冒泡排序对数字进行排序。

我的代码适用于80％的测试用例，但我找不到错误。我没有考虑到哪里或哪种情况？预先感谢。

#include <iostream>
#include <cmath>
using namespace std;

typedef long long int ll;

ll len(ll n){

    ll l = 0;
    while (n > 0){

        n /= 10;
        l++;
    }
return l;
}

ll concade(ll a,ll b){

    ll con = a * pow(10,len(b)) + b;
return con;
}


void swap(ll *xp,ll *yp){

    ll temp = *xp;
    *xp = *yp;
    *yp = temp;
}

void bubble_sort(ll arr[],ll n){

    ll i,j;
    for (i = 0; i < n - 1; i++){
        for (j = 0; j < n-i-1; j++){

            if (arr[j] == 0 && arr[j+1] != 0)
                swap(arr[j],arr[j+1]);

            if (concade(arr[j],arr[j+1]) < concade(arr[j+1],arr[j]))
                swap(arr[j],arr[j+1]);
        }
    }
}

int main(){

    ll n;
    while (cin >> n){

        if (n == 0) 
            break;

        ll arr[1000000];
        for(ll i = 0; i < n; i++){

            ll x;
            cin >> x;
            arr[i] = x;
        }
        bubble_sort(arr,n);
        for (ll i = 0; i < n; i++)
            cout << arr[i];
        cout << endl;
    }
}

解决方法

贪婪的方法可以很好地解决您的问题。

1. 根据第一个不同的数字对数字进行排序。
2. 将它们组合（按降序排列）

这是C＃中的实现（语法相似）

注意：您可以将字符串用于大型数据集，数字可用于小型数据集。并且考虑使用更快的算法进行排序，而不是气泡排序

 { "createGroup"               : [],"createGroupIfNotExistsFor" : ["groupMapper"],"deleteGroup"               : ["groupID"],"listPads"                  : ["groupID"],"createPad"                 : ["padID","text"],"createGroupPad"            : ["groupID","padName","createAuthor"              : ["name"],"createAuthorIfNotExistsFor": ["authorMapper","name"],"listPadsOfAuthor"          : ["authorID"],"createSession"             : ["groupID","authorID","validUntil"],"deleteSession"             : ["sessionID"],"getSessionInfo"            : ["sessionID"],"listSessionsOfGroup"       : ["groupID"],"listSessionsOfAuthor"      : ["authorID"],"getText"                   : ["padID","rev"],"setText"                   : ["padID","getHTML"                   : ["padID","setHTML"                   : ["padID","html"],"getRevisionsCount"         : ["padID"],"getLastEdited"             : ["padID"],"deletePad"                 : ["padID"],"getReadOnlyID"             : ["padID"],"setPublicStatus"           : ["padID","publicStatus"],"getPublicStatus"           : ["padID"],"listAuthorsOfPad"          : ["padID"],"padUsersCount"             : ["padID"]

这是使用C＃特殊语法的解决方案：

private static long F(List<int> numbers)
    {
        // 1
        for (int i = 0; i < numbers.Count; i++)
        {
            for (int j = 0; j < numbers.Count; j++)
            {
                if (Compare(numbers[i],numbers[j]) > 0)
                {
                    var temp = numbers[i];
                    numbers[i] = numbers[j];
                    numbers[j] = temp;
                }
            }
        }

        // 2
        long bigNumber = 0;
        for (int i = 0; i < numbers.Count; i++)
        {
            int numberOfDigits = NumberOfDigits(numbers[i]);
            bigNumber = (bigNumber * Pow10(numberOfDigits));
            bigNumber += numbers[i];
        }

        return bigNumber;
    }

    private static long Compare(long n1,long n2)
    {
        while (n1 > 0 && n2 > 0)
        {
            var difference = FirstDigit(n1) - FirstDigit(n2);
            if (difference != 0)
            {
                return difference;
            }
            n1 %= Pow10(NumberOfDigits(n1) - 1);
            n2 %= Pow10(NumberOfDigits(n2) - 1);
        }
        return FirstDigit(n1) - FirstDigit(n2);
    }

    private static long Pow10(long times)
    {
        long s = 1;
        for (int i = 0; i < times; i++)
        {
            s *= 10;
        }
        return s;
    }

    private static long FirstDigit(long n)
    {
        while (n >= 10)
        {
            n /= 10;
        }

        return n;
    }

    private static int NumberOfDigits(long n)
    {
        int digits = 1;
        while (n >= 10)
        {
            digits++;
            n /= 10;
        }

        return digits;
    }