Largest Number

Total Accepted: 35377 Total Submissions: 202492 Difficulty: Medium

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.

关键在于定义一种新的字符串比较规则，使得我们把排序好的字符串数组串一遍就能得到结果。

新的比较规则是return  s1+s2 < s2+s1;

然后证明这种比较规则为什么是对的。

证明：
假设按这种规则排序好的数组是：A1，A2，A3...Ax....Ay.....An

根据这种排序规则，最后的排序结果是An...AyAy-1...Ax...A3A2A1

反证法证明：

假设存在一个An...AxAy-1...Ay...A3A2A1 > An...AyAy-1...Ax...A3A2A1

由于Ax<Ax+1<Ax+2<Ax+3<....<Ay-1<Ay 

那么An...Ay Ay-1... Ax+2 Ax+1 Ax...A3 A2 A1 > An...AyAy-1...Ax Ax+1...A3A2A1，这是根据比较规则来的，比较规则中，若Ax<Ax+1则AxAx+1 < Ax+1Ax

同样An...AyAy-1...Ax Ax+1...A3A2A1 > An...AyAy-1...Ax Ax+2 Ax+1...A3A2A1>...>An...Ax Ay Ay-1...Ax+2 Ax+1...A3A2A1。

同理可以证明An...Ax Ay Ay-1...Ax+2 Ax+1...A3A2A1>An...Ax Ay-1...Ax+1 Ay...A3A2A1。

这就与假设相反，所以得证

bool comp(const string& lhs,const string& rhs){
    return lhs+rhs < rhs+lhs;
}
class Solution {
public:
    string largestNumber(vector<int>& nums) {
        vector<string> str_nums;
        //以下三种整数转字符串
        /*1.
        for(auto num:nums){
            char buf[20]={0};
            sprintf(buf,"%d",num);
            str_nums.push_back(string(buf));
        }
        */
        /*2.
        vector<string> str_nums(nums.size());
        stringsteam stream;
        for(auto num:nums){
            stream<<num;
            stream>>str_nums[i];
            str_nums.push_back(string(buf));
        }
        */
        for(auto num:nums){
            str_nums.push_back(to_string(num));
        }
        sort(str_nums.begin(),str_nums.end(),comp);
        string res;
        for(int i=str_nums.size()-1;i>=0;i--){
            res.append(str_nums[i].begin(),str_nums[i].end());
        }
        return (res.empty() || res[0]!='0' )?  res : "0";
    }
};