leetcode-Longest Valid Parentheses
2014-09-25 16:56
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Given a string containing just the characters
find the length of the longest valid (well-formed) parentheses substring.
For
which has length = 2.
Another example is
is
class Solution {
public:
int longestValidParentheses(string s) {
int n = 0;
int ret = 0;
int i = 0;
vector<int> a;
vector<int> b;
while(s[i])
{
int m = a.size();
if(n == 0)
{
if(s[i] =='(')
{
a.push_back(1);
b.push_back(0);
n++;
}
else if(s[i] == ')')
{
a.push_back(-1);
b.push_back(0);
}
}
else if(n > 0)
{
if(s[i] == '(')
{
if(s[i-1] == ')')
{
a.push_back(1);
b.push_back(0);
n++;
}
else if(s[i-1] == '(')
{
n++;
a[m-1]++;
}
}
else if(s[i] == ')')
{
while(a[m-1] == 0)m--;
a[m-1]--;
b[m-1]++;
n--;
}
}
i++;
}
i = 0;
int m = a.size();
while(i < m-1)
{
int count = b[i];
while ((i < m-1)&&(a[i+1] == 0))
{
count += b[++i];
}
if(count > ret)ret = count;
i++;
}
if((m > 0)&&(b[m-1] > ret))ret = b[m-1];
return ret*2;
}
};
'('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.
class Solution {
public:
int longestValidParentheses(string s) {
int n = 0;
int ret = 0;
int i = 0;
vector<int> a;
vector<int> b;
while(s[i])
{
int m = a.size();
if(n == 0)
{
if(s[i] =='(')
{
a.push_back(1);
b.push_back(0);
n++;
}
else if(s[i] == ')')
{
a.push_back(-1);
b.push_back(0);
}
}
else if(n > 0)
{
if(s[i] == '(')
{
if(s[i-1] == ')')
{
a.push_back(1);
b.push_back(0);
n++;
}
else if(s[i-1] == '(')
{
n++;
a[m-1]++;
}
}
else if(s[i] == ')')
{
while(a[m-1] == 0)m--;
a[m-1]--;
b[m-1]++;
n--;
}
}
i++;
}
i = 0;
int m = a.size();
while(i < m-1)
{
int count = b[i];
while ((i < m-1)&&(a[i+1] == 0))
{
count += b[++i];
}
if(count > ret)ret = count;
i++;
}
if((m > 0)&&(b[m-1] > ret))ret = b[m-1];
return ret*2;
}
};
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