题目描述

Since the answer may not fit in an integer data type, return the answer as a string.

If there is no answer return an empty string.

Example 1:

Input: digits = [8,1,9]
Output: "981"

Example 2:

Input: digits = [8,6,7,1,0]
Output: "8760"

Example 3:

Input: digits = [1]
Output: ""

Example 4:

Input: digits = [0,0,0,0,0,0]
Output: "0"

Constraints:

• 1 <= digits.length <= 10^4
• 0 <= digits[i] <= 9
• The returning answer must not contain unnecessary leading zeros.

分析

能被3整除的数各位数的和sum必能被3整除，反之也成立。因此，计算各位之和，计算res=sum%3，若res!=0，删掉最小的数使其满足。容易忽视一点：若res == 1，不仅可以删掉一个除3余1的数，也可以删掉两个除3余2的数。对于res == 2也是如此。

AC代码

这里模仿的评论区的一位大神的思路写的：

class Solution {
    int cnt[10]={0},sum=0;
public:
    bool Delete(int n){
        for(int i=n;i<=9;i+=3) if(cnt[i]>0) {cnt[i]--;return 1;}
        return 0;
    }
    string largestMultipleOfThree(vector<int>& digits) {
        string ans;
        for(int k:digits) {sum+=k;cnt[k]++;}
        if(sum%3==1) if(!Delete(1)) Delete(2),Delete(2);
        if(sum%3==2) if(!Delete(2)) Delete(1),Delete(1);
        for(int i=9;i>=0;i--) while(cnt[i]--) ans+=i+'0';
        if(ans.size()>0&&ans[0]=='0') return "0";
        return ans;
    }
};