[LeetCode] Valid Parentheses、Generate Parentheses、Longest Valid Parentheses

这三道题是大摩实习现场面试时候的笔试题。

Valid Parentheses：

Given a string containing just the characters

'('

,

')'

,

'{'

,

'}'

,

'['

and

']'

,
determine if the input string is valid.

The brackets must close in the correct order,

"()"

and

"()[]{}"

are
all valid but

"(]"

and

"([)]"

are
not.

比较简单的栈操作，但是居然还是发现有问题。一个是switch语句不会写 - -， 二个是又是拉掉一些小细节，比如匹配了之后应当出栈，又忘了，最后要判断为空也忘了，哎，算起来100分的话也就能得30分了。。。。。

class Solution {
public:
bool isValid(string s) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

int n=s.length();
stack<char> st;
for(int i=0;i<n;i++)
{
switch(s[i])
{
case '(':
case '[':
case '{':
st.push(s[i]);
break;
case ')':
if ( st.empty() || st.top()!='(')
return false;
st.pop();
break;
case ']':
if ( st.empty() || st.top()!='[')
return false;
st.pop();
break;
case '}':
if ( st.empty() || st.top()!='{')
return false;
st.pop();
break;
default:
return false;
}
}
return st.empty();
}
};

Generate Parentheses:

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))",
"(()())", "(())()", "()(())", "()()()"

写的时候又犯了好几个错误，哎。尤其是rec(n+1)这里老是忘了+1.

class Solution {
public:
vector<string> generateParenthesis(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
rec=vector<vector<string> >(n+1);
solve(n );
return rec
;
}
vector<vector<string> > rec;
void solve(int n)
{
if ( rec
.size()!=0 )
return;
if ( n==0)
{
rec[0]=vector<string> (1,string());
return;
}
else if ( n==1 )
{
rec[1]=vector<string> (1,string("()"));
return;
}
else
{
for(int k=0;k<n;k++)
{
solve(k);
solve(n-1-k);
vector<string>& in=rec[k];
vector<string>& out=rec[n-1-k];
int inNum=rec[k].size();
int outNum=rec[n-1-k].size();
string tmp;
for(int i=0;i<inNum;i++)
for(int j=0;j<outNum;j++)
{
tmp.clear();
tmp.push_back('(');
tmp.append(in[i]);
tmp.push_back(')');
tmp.append(out[j]);
rec
.push_back(tmp);
}
}
}
}
};

Longest Valid Parentheses：

Given a string containing just the characters

'('

and

')'

,
find the length of the longest valid (well-formed) parentheses substring.

For

"(()"

,
the longest valid parentheses substring is

"()"

,
which has length = 2.

Another example is

")()())"

,
where the longest valid parentheses substring is

"()()"

,
which has length = 4.
两种方法，一种是利用栈：

class Solution {
public:
int longestValidParentheses(string s) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

typedef pair<char,int> record;
stack<record> st;
st.push(record('#',-1));
int ans=0;
int n=s.length();
for(int i=0;i<n;i++)
if ( s[i]==')'&&st.top().first=='(')
st.pop();
else
{
ans=max(ans,i-st.top().second-1);
st.push(record(s[i],i));
}
ans=max(ans,n-st.top().second-1);
return ans;
}
};

第二种方法类似DP，思路见/article/8613984.html：

class Solution {
public:
int longestValidParentheses(string s) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

int n=s.length();
vector<int> right(n,-1);
for(int i=n-1;i>=0;i--)
{
if( s[i]=='(')
{
int j=i+1;
while( j<n&&right[j]!=-1)
j=right[j]+1;
if ( j<n && s[j]==')')
right[i]=j;
}
}
int ans=0;
for(int i=0;i<n;i++)
if ( right[i]!=-1)
{
int j=right[i]+1;
while(j<n&&right[j]!=-1)
j=right[j]+1;
ans=max(ans,j-i);
}
return ans;
}
};