其实大部分是东京时间第二天凌晨才做的- -但国际服刷新比较晚
BGM:刀剑如梦
Decsripition
Given a string s of '(' , ')' and lowercase English characters.
Your task is to remove the minimum number of parentheses ( '(' or ')', in any positions ) so that the resulting parentheses string is valid and return any valid string.
Formally, a parentheses string is valid if and only if:
- It is the empty string, contains only lowercase characters, or
- It can be written as AB (A concatenated with B), where A and B are valid strings, or
- It can be written as (A), where A is a valid string.
Example
Example 1:
Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.
Example 2:
Input: s = "a)b(c)d"
Output: "ab(c)d"
Example 3:
Input: s = "))(("
Output: ""
Explanation: An empty string is also valid.
Solution
Just use stack to pair left and right parentheses, but remeber to save the num-th of the remain char.
Code
cpp
class Solution {
public:
string minRemoveToMakeValid(string s)
{
typedef struct pair
{
char kyara;
int num;
};
stack<pair>tree;
bool judge[100005]={0};
for(int i=0;i<s.length();++i)
{
if(s[i]=='(')
{
pair mid;
mid.kyara=s[i];
mid.num=i;
tree.push(mid);
}
else if(s[i]==')')
{
if(!tree.empty())
{
pair mid=tree.top();
if(mid.kyara=='(') tree.pop();
else
{
pair curr;
curr.kyara=s[i];
curr.num=i;
tree.push(curr);
}
}
else
{
pair mid;
mid.kyara=s[i];
mid.num=i;
tree.push(mid);
}
}
}
while(!tree.empty())
{
pair mid=tree.top();
judge[mid.num]=1;
tree.pop();
}
string c="";
for(int i=0;i<s.length();++i)
{
if(judge[i]==0) c.push_back(s[i]);
}
return c;
}
};