一开始考虑出错，主要是因为它有一个子串的限制，就是要求子串是连续的。返回返回长度最小的子串长度。这个是错误的解法。
class Solution {
public:int balancedString(string s) {vector<int> cnt(4, 0);for(int i=0;i<s.size();i++){if (s[i]=='Q'){cnt[0]++;}if (s[i]=='W'){cnt[1]++;}if (s[i]=='E'){cnt[2]++;}if (s[i]=='R'){cnt[3]++;}}int target = int(s.size()/4);int res = 0;for(int i=0;i<4;i++){cout<<cnt[i]<<endl;if (cnt[i]>target){res += cnt[i]-target;}}return res;}
};
改成子串处理，对是对了，但是超时了，38/40
class Solution {
public:int balancedString(string s) {// gain all numbersmap<char, int> cnt;for(int i=0;i<s.size();i++){if (cnt.find(s[i])!=cnt.end()){cnt[s[i]]++;}else{cnt[s[i]] = 1;}}int target = int(s.size()/4);//if original is okif (cnt['Q']<=target&&cnt['W']<=target&&cnt['E']<=target&&cnt['R']<=target){return 0;}// original is not ok, remove windows [i, j] and judge againint res = s.size();for(int i=0;i<s.size();i++){map<char, int> cntnow = cnt;for(int j=i;j<s.size();j++){cntnow[s[j]] --;if (cntnow['Q']<=target&&cntnow['W']<=target&&cntnow['E']<=target&&cntnow['R']<=target){res = res < (j-i+1)? res: (j-i+1);break;}}}return res;}
};
看了别人的，真是学不来啊，佩服佩服
class Solution {
public:int balancedString(string s) {unordered_map<int, int> count;int n = s.length(), res = n, i = 0, k = n / 4;for (int j = 0; j < n; ++j) {count[s[j]]++;}for (int j = 0; j < n; ++j) {count[s[j]]--;while (i < n && count['Q'] <= k && count['W'] <= k && count['E'] <= k && count['R'] <= k) {res = min(res, j - i + 1);count[s[i++]] += 1;}}return res;}
};