301. Remove Invalid Parentheses(bfs，hard)

dfs 不易处理，易超时
主要是借助队列和set来做(相当于一层一层的来找，找到就可以退出，因为要求出最小删除)

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题目地址

https://leetcode.com/problems/remove-invalid-parentheses/

题目描述

Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.

Note: The input string may contain letters other than the parentheses ( and ).
Examples:

"()())()" -> ["()()()", "(())()"]
"(a)())()" -> ["(a)()()", "(a())()"]
")(" -> [""]

ac

class Solution {
public:

    //判断是否是合法的字符串
    bool isValid(string s)
    {
        int count = 0;
        for (int i = 0; i<s.length(); i++)
        {
            if (s[i] == '(')
                count++;
            else if (s[i] == ')'){
                if (count == 0)
                    return false;
                count--;
            }
        }
        return count == 0;
    }

    vector<string> ret;

    vector<string> removeInvalidParentheses(string s)
    {
        unordered_map<string, int> map;
        queue<string> q;
        q.push(s);
        map[s] = 1;
        bool found = false;
        while (!q.empty())
        {
            string str = q.front();
            q.pop();
            if (isValid(str)){
                ret.push_back(str);
                found = true;
            }
            if (found)
                continue;
            for (int i = 0; i<str.length(); i++)
            {
                if (str[i] != ')'&&str[i] != '(')
                    continue;
                //将这个括号从字符串中去除, 即这个括号要或者不要
                string sub = str.substr(0, i) + str.substr(i + 1);
                if (map.find(sub) == map.end()) //避免重复
                {
                    q.push(sub);
                    map[sub] = 1;
                }
            }
        }
        return ret;
    }
};