LeetCode Remove Invalid Parenthese

LeetCode 每日刷题！！

Remove Invalid Parentheses

这道题在做的时候用到了BFS因为它要求的是最优解

层次依次为删除0个元素，1个元素，2个元素。。。

如果在一层找到符合要求的就返回

在做的时候考虑效率问题，要对BFS进行剪枝，不然会timelimit，使用的方法一：出现过的就加入visited里面不再加入叶子节点，二：visited用hash，提高查找效率

#include <iostream>
#include <stdio.h>
#include <string>
#include <map>
#include <vector>
#include <algorithm>
#include <unordered_set>
#include <queue>
#include <stack>
#include <assert.h>
#include <set>
using namespace std;class Solution {
public:vector<string> removeInvalidParentheses(string s) {set<string> visited;string *current = &s;vector<string> ans;vector<string> layerSet;vector<string> nextlayer;if (s == ""){ans.push_back("");return ans;}layerSet.push_back(s);while (layerSet.size() != 0){//each layer of a BFT, the optimize have the least delete number//and each loop have the same delete num.for (int i = 0; i < layerSet.size(); i++){*current = layerSet[i];if (isValid(*current)){ans.push_back(*current);}}if (ans.size() != 0){return ans;}for (int i = 0; i < layerSet.size(); i++){*current = layerSet[i];for (int i = 0; i < current->size(); i++){if ((*current)[i] != '(' && (*current)[i] != ')'){continue;}string tmp = current->substr(0, i) + current->substr(i + 1);if (tmp == ""){ans.push_back("");return ans;}if (visited.find(tmp) == visited.end()){nextlayer.push_back(tmp);visited.insert(tmp);}}}layerSet.swap(nextlayer);}return ans;}bool isValid(string s){int num = 0;if (s.size() == 0){return false;}for (int i = 0; i < s.size(); i++){if (s[i] == '(') {num++;}if (s[i] == ')') {num--;}if (num < 0)return false;}return num == 0;}
};int main()
{string test1 = ")()m)(((()((()((((";vector<string> ans;Solution *solu = new Solution();ans = solu->removeInvalidParentheses(test1);for (int i = 0; i < ans.size(); i++){cout << ans[i] << endl;}system("pause");return 0;}

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