leetcode problem 32 -- Longest Valid Parentheses

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.

思路1：

　　从左到右依次扫描，用一个数组v记录括号是否已经匹配。然后再扫描一次v，用一个变量count记录连续合法括号数目。如果v[i] = 0则说明该括号未匹配，合法括号序列终端，count = 0. 代码如下：

Runtime: 22 ms

class Solution {
public:
int longestValidParentheses(string s) {
vector<int> v(s.length(), 0);
stack<int> stac;
for (int i = 0; i < s.length(); ++i) {
if (s[i] == ')') {
if (stac.empty())
continue;
else {
v[i] = 2;
v[stac.top()] = 2;
stac.pop();
}
}
else
stac.push(i);
}

int res = 0, count = 0;
for (int i = 0; i < s.length(); ++i) {
if (v[i] == 0) {
if (res < count)
res = count;
count = 0;
}
else if (s[i] == '(')
count += v[i];
}
return res < count ? count : res;
}

};

思路二：

　　先从左到右扫描，分别记录'('和')'的数目， 如果一旦')'的数目大于'(' 那么可以确定合法连续括号序列中断。记录countMax。

　　然后从右往左扫描，同样分别记录'('和')'的数目， 如果一旦'('的数目大于')' , 那么可以确定合法连续括号序列中断。记录countMax。

代码如下：

Runtime: 10 ms

class Solution {
public:
int longestValidParentheses(string s) {
int ll = 0, lr = 0, li = 0;
int rl = 0, rr = 0, ri = s.length()-1;
int res = 0;
for (; li < s.length() && ri >= 0; ++li, --ri) {
switch (s[li]) {
case '(':
++ll;
break;
case ')':
++lr;
}
switch (s[ri]) {
case '(':
++rl;
break;
case ')':
++rr;
}
if (ll == lr && (ll * 2) > res)
res = 2 * ll;
else if (ll < lr)
ll = lr = 0;

if (rl == rr && (rl * 2) > res)
res = 2 * rl;
else if (rl > rr)
rl = rr = 0;

}

return res;
}
private:

};