删除最小数目的无效括号，使输入的字符串有效，返回所有可能的结果。
注意: 输入可能包含了除 ( 和 ) 以外的元素。
示例 :
"()())()" -> ["()()()", "(())()"]
"(a)())()" -> ["(a)()()", "(a())()"]
")(" -> [""]
详见：https://leetcode.com/problems/remove-invalid-parentheses/description/
方法一：
class Solution {
public: vector removeInvalidParentheses(string s) { vector ans; helperDFS(s, ')', 0,ans); return ans; } void helperDFS(string s, char ch, int last,vector &ans) { for(int i = 0, cnt = 0; i < s.size(); i++) { if(s[i]=='('||s[i]==')'){s[i]==ch?cnt++:cnt--; }if(cnt <= 0){continue; }for(int j = last; j <= i; j++) { if(s[j] == ch && (j ==last || s[j-1]!= ch)) {helperDFS(s.substr(0, j)+s.substr(j+1), ch, j,ans); }} return; } reverse(s.begin(), s.end()); if(ch == ')'){return helperDFS(s, '(', 0,ans); }ans.push_back(s); }
};
方法二：
class Solution {
public:vector removeInvalidParentheses(string s) {vector res;unordered_set visited{ {s}};queue q{ {s}};bool found = false;while (!q.empty()){string t = q.front();q.pop();if (isValid(t)) {res.push_back(t);found = true;}if (found){continue;}for (int i = 0; i < t.size(); ++i) {if (t[i] != '(' && t[i] != ')'){continue;}string str = t.substr(0, i) + t.substr(i + 1);if (!visited.count(str)) {q.push(str);visited.insert(str);}}}return res;}bool isValid(string t) {int cnt = 0;for (int i = 0; i < t.size(); ++i) {if (t[i] == '('){++cnt;}else if (t[i] == ')' && --cnt < 0){return false;}}return cnt == 0;}
};
参考：https://blog.csdn.net/qq508618087/article/details/50408894
https://www.cnblogs.com/grandyang/p/4944875.html