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：当碰到")"时，向左遍历找到未标记过的"("，则将这两个位置标为1，然后统计1连续出现的长度即可，时间复杂度为O(n^2)。
class Solution {
public:
int longestValidParentheses(string s) {
int n = s.length();
if (n<1) {
return 0;
}
bool matched[n];
memset(matched,0,sizeof(matched));
int i,j;
for(i=0; i<n; ++i) {
if(s[i] == ')') {
j = i - 1;
while(j>=0) {
if(!matched[j]&&s[j]=='(') {
matched[j] = true;
matched[i] = true;
break;
}
--j;
}
}
}
int co = 0;
int max = INT_MIN;
for(i=0; i<n; ++i) {
if(matched[i]) {
co++;
}
else {
max = (max<co?co:max);
co = 0;
}
}
if (co>0) {
max = (max<co?co:max);
}
return (max==INT_MIN?0:max);
}
};