ARST打卡第337周

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class Solution {
public:
    string findLexSmallestString(string s, int a, int b) {
        // 按照上面思路实现是可行的，这里直接实现
        int n = s.size();
        string ans = s;
        unordered_set<string> vis;
        queue<string> q;
        q.push(s);
        vis.insert(s);
        while (!q.empty()) {
            string cur = q.front(); q.pop();
            if (cur < ans) ans = cur;
            // 操作1：给奇数下标加a
            string t = cur;
            for (int i = 1; i < n; i += 2) {
                t[i] = ((t[i] - '0' + a) % 10) + '0';
            }
            if (vis.find(t) == vis.end()) {
                vis.insert(t);
                q.push(t);
            }
            // 操作2：旋转b位
            string t2 = cur.substr(n - b) + cur.substr(0, n - b);
            if (vis.find(t2) == vis.end()) {
                vis.insert(t2);
                q.push(t2);
            }
        }
        return ans;
    }
};