32. Longest Valid Parentheses
Problem‘s Link
----------------------------------------------------------------------------
Mean:
给定一个由‘(‘和‘)‘组成的字符串，求最长连续匹配子串长度.
analyse:
定义一个stack<pair<char,int>>类型的stack.
遇到‘(‘进栈;
遇到‘)‘需要分两种情况讨论：
- 栈顶元素为‘(‘，正好匹配，将栈顶元素出栈;
- 栈顶元素为‘)‘或栈为空，将当前符号和下标入栈.
这个栈构建完以后，我们只需要找这个栈中相邻两个top.second()之差即可.
trick：注意‘(()‘这种情况，需要特判，即backVal的值为len.
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-01-20.35
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
class Solution
{
public:
int longestValidParentheses(string s)
{
int len=s.length();
stack<pair<char,int>> sta;
for(int i=0;i<len;++i)
{
if(s[i]==‘(‘)
sta.push(make_pair(‘(‘,i));
else
{
if(!sta.empty())
{
pair<char,int> t=sta.top();
if(t.first==‘(‘)
sta.pop();
else
sta.push(make_pair(‘)‘,i));
}
else
sta.push(make_pair(‘)‘,i));
}
}
int ans=0,frontVal=len,backVal;
if(!sta.empty() &&sta.top().second!=len-1)
backVal=len;
else
backVal=len-1;
while(!sta.empty())
{
pair<char,int> t=sta.top();
sta.pop();
frontVal=t.second;
ans=max(ans,backVal-frontVal-1);
backVal=frontVal;
}
ans=max(ans,frontVal-0);
return ans;
}
};
int main()
{
Solution solution;
string s;
while(cin>>s)
{
cout<<solution.longestValidParentheses(s)<<endl;
}
return 0;
}
/*
*/