1347. Minimum Number of Steps to

思路：step1：用两个map分别存储两个串字符重复的个数；
step2：比较两个key相同的值的差异，如果map_s[k] > map_t[k]，叠加；
代码：

class Solution {
public:
    map <char, int> getMap(string str) {
        map<char, int> res;
        for (int i = 0; i < str.length(); i++) {
            if (res.count(str[i])) res[str[i]] ++;
            else res[str[i]] = 1;
        }
        return res;
    } 
    int minSteps(string s, string t) {
        int res = 0;
        map <char, int> map_s = getMap(s);
        map <char, int> map_t = getMap(t);
        map <char, int>::iterator iter = map_s.begin();
        while(iter != map_s.end()) {
            map_t.count(iter -> first) ?: map_t[iter -> first] = 0;
            cout << map_t[iter -> first] << iter -> second;
            if (iter -> second > map_t[iter -> first]) res += iter -> second - map_t[iter -> first];
            iter ++;
        }
        return res;
    }
};