179. Largest Number

这题有很好的解法，也有自己想出来的解法。
完美解法：

class Solution {
public:
    string largestNumber(vector<int>& nums) {
        vector<string> strVec;
        for (int num : nums)
            strVec.push_back(to_string(num));
        sort(strVec.begin(), strVec.end(), sortHelper);
        stringstream ss;
        for (string& str : strVec)
            ss << str;
        string retStr = ss.str();
        int idx = 0;
        while (idx < retStr.size() && retStr[idx] == '0')
            ++idx;
        return idx == retStr.size()? "0" : retStr.substr(idx);
    }
private:
    
    static bool sortHelper(string& num1, string& num2) {
        return num1+num2 > num2+num1;
    }
};

自己的解法：

class Solution {
public:
    string largestNumber(vector<int>& nums) {
        vector<string> strVec;
        for (int num : nums)
            strVec.push_back(to_string(num));
        sort(strVec.begin(), strVec.end(), sortHelper);
        stringstream ss;
        for (string& str : strVec)
            ss << str;
        string retStr = ss.str();
        int idx = 0;
        while (idx < retStr.size() && retStr[idx] == '0')
            ++idx;
        return idx == retStr.size()? "0" : retStr.substr(idx);
    }
private:
    static bool sortHelper(string num1, string num2) {
        int idx1 = 0, idx2 = 0;
        while (idx1 < num1.size() && idx2 < num2.size() && num1[idx1] == num2[idx2]) {
            ++idx1;
            ++idx2;
        }
        if (idx1 < num1.size() && idx2 < num2.size())
            return num1[idx1] > num2[idx2];
        if (idx1 == num1.size() && idx2 == num2.size())
            return true;
        if (idx1 == num1.size())
            return sortHelper(num1, num2.substr(idx2));
        else
            return sortHelper(num1.substr(idx1), num2);
    }
};

第一次写在sorthelper当中调用自身的。虽然这份代码有一个例子没有通过，但是示例都通过了。之后可以再修改修改