leetcode 5232 Replace the Substring for Balanced String

https://leetcode-cn.com/problems/replace-the-substring-for-balanced-string/

滑动窗口+判定。对于窗口大小为len。窗口外的每个字母出现次数小于等于n/4时候，一定可以替换成功，这时就可以尝试更小的窗口。对于窗口大小的选择，我刚开始用二分，O(nlgn)的复杂度，结果 n=10^5 会超时。后来发现双指针就可以O(n)就过了。

二分：

class Solution {
public:
    char chs[4] = {'Q', 'W', 'E', 'R'};
    unordered_map<char, int> cnt;
    bool judge(string s, int len)
    {
        int n = s.length();
        for(int i = 0; i + len - 1 < n; i++)
        {
            unordered_map<char, int> tmp = cnt;
            for(int j = i; j < i + len; j++)
                tmp[s[j]]--;
            int need = 0;
            for(int j = 0; j < 4; j++)
            {
                if(tmp[chs[j]] > n / 4) need = n;
                else need += n / 4 - tmp[chs[j]];
            }
            if(len >= need) return true;
        }
        return false;
    }
    int balancedString(string s) {
        int n = s.length();
        int l = 0, r = n;
        for(char ch : s)
            cnt[ch] ++;
        while(l < r)
        {
            int mid = l + r >> 1;
            if(judge(s, mid)) r = mid;
            else l = mid + 1;
        }
        return l;
    }
};

双指针：

class Solution {
public:
    int balancedString(string s) {
        unordered_map<char, int> cnt;
        for(char ch : s)
            cnt[ch]++;
        int l = 0, r = 0;
        int n = s.length(), need = n / 4; 
        int res = n;
        while(r <= n)
        {
            if(cnt['Q'] > need || cnt['W'] > need || cnt['E'] > need || cnt['R'] > need)
            {
                cnt[s[r++]]--;
                continue;
            }
            res = min(res, r - l);
            if(res == 0) break;
            cnt[s[l++]]++;
        }
        return res;
    }
};