Largest Number

最新推荐文章于 2021-02-17 06:13:28 发布

原创最新推荐文章于 2021-02-17 06:13:28 发布 · 359 阅读

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本文介绍了一种算法，用于将非负整数列表重新排列形成最大的可能数值。通过比较字符串的不同组合来确定最佳顺序，并讨论了实现过程中的关键挑战。

Given a list of non negative integers, arrange them such that they form the largest number.

For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330.

Note: The result may be very large, so you need to return a string instead of an integer.

一开始脑子没转过弯来，用的复杂的判断大小方法，一直通不过各种 test case

class Solution {
public:
    void execute(string &str, vector<string> &strnum)
    {
        while(strnum.begin() != strnum.end())
        {
            vector<string>::iterator big = strnum.begin();
            vector<string>::iterator iter = big+1;
            while (iter != strnum.end())
            {
                if (big->compare(*iter) == 0)
                 {
                     iter++;
                     continue;
                 }
                int i = 0;
                while(1)
                {
                    if (i == big->size())
                     {
                         if (big->compare(iter->substr(i)) < 0)
                             big = iter;
                        break;
                     }
                    if (i == iter->size())
                    {
                       if (iter->compare(big->substr(i)) > 0)
                            big = iter;
                        break;
                    }
                    if ((*big)[i] > (*iter)[i])
                        break;
                    if ((*big)[i] < (*iter)[i])
                    {
                        big = iter;
                        break;
                    }
                    i++;
                }
                iter++;
            }
            str += *big;
            strnum.erase(big);
        }
    }
    string largestNumber(vector<int> &num) 
    {
        string res;
        vector< vector<string> > strnum(10);
        for (int i = 0; i < num.size(); i++)    
         {
             stringstream ss;
             ss << num[i];
             string tmp = ss.str();
             strnum[tmp[0] - '0'].push_back(tmp);
         }
        
         for (int i = 9; i > 0; i--)
           execute(res, strnum[i]);
         if (res.empty())
            return "0";
        execute(res, strnum[0]);
         return res;
    }
};

后来不知怎么脑袋就开窍了就通过了，难点就在于判断俩首字母相同的字符串大小吧

class Solution {
public:
    void execute(string &str, vector<string> &strnum)
    {
        while(strnum.begin() != strnum.end())
        {
            vector<string>::iterator big = strnum.begin();
            vector<string>::iterator iter = big+1;
            while (iter != strnum.end())
            {
                string tmp1 = *big;
                tmp1 += *iter;
                string tmp2 = *iter;
                tmp2 += *big;
                if (tmp1.compare(tmp2) < 0)
                   big = iter;
                 iter++;
            }
            str += *big;
            strnum.erase(big);
        }
    }
    string largestNumber(vector<int> &num) 
    {
        string res;
        vector< vector<string> > strnum(10);
        for (int i = 0; i < num.size(); i++)    
         {
             stringstream ss;
             ss << num[i];
             string tmp = ss.str();
             strnum[tmp[0] - '0'].push_back(tmp);
         }
        
         for (int i = 9; i > 0; i--)
           execute(res, strnum[i]);
           
         if (res.empty())
            return "0";
            
         if(strnum[0].size() > 0)
         {
             string tmp(strnum[0].size(), '0');
             res += tmp;
         }
        return res;
    }
};