题目
方法一:dp
int longestValidParentheses(string s) {
int ans = 0
stack<int> sta
sta.push(-1)
for(int i = 0
{
if(s[i] == '(')sta.push(i)
else{
sta.pop()
if(sta.empty())
{
sta.push(i)
}else{
ans = max(ans,i - sta.top())
}
}
}
return ans
}
方法二:栈
int longestValidParentheses(string s) {
s = ' ' + s
int n = s.size()
vector<int> f(n,0)
for(int i = 2
{
if(s[i] == ')'){
if(s[i-1] == '(')f[i] = f[i-2] + 2
else if(s[i - f[i-1] - 1] == '(')f[i] = f[i - f[i-1] - 2] + f[i-1] + 2
}
}
return *max_element(f.begin(),f.end())
}
方法三:贪心
int longestValidParentheses(string s) {
int left = 0,right = 0
int ans = 0
for(int i = 0
{
if(s[i] == '(')left ++
else right ++
if(left == right) ans = max(ans,left * 2)
else if(right > left)right = left = 0
}
left = 0,right = 0
for(int i = s.size() - 1
{
if(s[i] == '(')left ++
else right ++
if(left == right) ans = max(ans,left * 2)
else if(right < left)right = left = 0
}
return ans
}
相关题目
void solve()
{
string s
int maxLen = longestValidParentheses(s)
int left = 0,right = 0
int maxNum = 1,num = 0
for(int i = 0
{
if(s[i] == '(')left ++
else right ++
if(right > left)right = left = 0
else if(right == left && right*2 == maxLen)num ++
}
maxNum = max(maxNum,num)
// debug1(num)
left = right = num = 0
for(int i = s.size() - 1
{
if(s[i] == '(')left ++
else right ++
if(right < left)right = left = 0
else if(right == left && right*2 == maxLen)num ++
}
// debug1(num)
maxNum = max(maxNum,num)
cout << maxLen << " " << maxNum << endl
}