Acwing74

标签：maxlen Acwing74 ++ int str ans size

k个k进制的数字进行不进位加法，结果为0

时空复杂度O(n)

class Solution {
public:
    string decTok(int dec,int k){
        string ret = "";
        while(dec){
            ret += char(dec%k + '0');
            dec /= k;
        }
        reverse(ret.begin(),ret.end());
        return ret;
    }
    
    int kTodec(string ans,int k){
        int times = ans.size() - 1;
        int res = 0;
        for(int i  = 0; i < ans.size();i++){
            res += (ans[i] - '0')* pow(k,times--);
        }
        return res;
    }
    int findNumberAppearingOnce(vector<int>& nums) {
        int size = nums.size(),k = 3,maxlen = 0;
        string str[size];
        for(int i = 0; i < size;i++){
            str[i] = decTok(nums[i],k);
        }
        for(int i = 0; i < size;i++){
            int len = str[i].size();
            maxlen = max(maxlen,len);
        }
        for(int i = 0; i < size; i++){
            while(str[i].size() < maxlen){
                str[i] = "0" + str[i];
            }
        }
        string ans(maxlen,'0');
        for(int i = 0;i < size;i++){
            for(int j = 0;j < maxlen;j++){
                ans[j] = char(((str[i][j] - '0') + (ans[j] - '0')) % k + '0');
            }
        }
        return kTodec(ans,k);
    }
};

标签：maxlen,Acwing74,++,int,str,ans,size
From： https://www.cnblogs.com/poteitoutou/p/17077298.html