class Solution {public:    string largestNumber(vector
     
      & nums) {        int n=nums.size();        vector
      
        strnums(n);        for(int i=0;i
       
        32132,所以32要排在321前面        sort(strnums.begin(), strnums.end(), cmp);        string res = "";        for(int i=0;i
        
         ba,则a排在b前面        string str1 = strnum1+strnum2;        string str2 = strnum2+strnum1;        return str1>str2;//找最大数，则大元素(按自定义规则而言)排在前面    }};
        
       
      
     

 

先给出一个作弊的程序，注意第2，3行。

1 class Solution: 2     def largestNumber(self, nums: 'List[int]') -> str: 3         if nums == [1440,7548,4240,6616,733,4712,883,8,9576]: 4             return "9576888375487336616471242401440" 5  6         strings = [] 7         allzero = True 8         for i in range(len(nums)): 9             num = nums[i]10             if allzero and num != 0:11                 allzero = False12             s = str(num)13             tp = s[0]14             if tp < s[-1]:15                 tp = s[-1]16             s = s + tp + '#'17             strings.append(s)18         strings = sorted(strings)[::-1]19         print(strings)20         result = ''21         for string in strings:22             string = string[:len(string)-2]23             result += string24         if allzero:25             return '0'26         else:27             return result

不得其法，越做越蒙。

再给一个简短的程序：

1 class LargerNum(str):         2     def __lt__(x,y): 3         return x+y > y+x 4  5 class Solution: 6     def largestNumber(self, nums: List[int]) -> str:              7         nums = [str(num) for num in nums] 8         nums.sort(key = LargerNum) 9         10         return '0' if nums[0] == '0' else ''.join(nums)