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(), );
stack<int> stac;
for (int i = ; i < s.length(); ++i) {
if (s[i] == ')') {
if (stac.empty())
continue;
else {
v[i] = ;
v[stac.top()] = ;
stac.pop();
}
}
else
stac.push(i);
} int res = , count = ;
for (int i = ; i < s.length(); ++i) {
if (v[i] == ) {
if (res < count)
res = count;
count = ;
}
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 = , lr = , li = ;
int rl = , rr = , ri = s.length()-;
int res = ;
for (; li < s.length() && ri >= ; ++li, --ri) {
switch (s[li]) {
case '(':
++ll;
break;
case ')':
++lr;
}
switch (s[ri]) {
case '(':
++rl;
break;
case ')':
++rr;
}
if (ll == lr && (ll * ) > res)
res = * ll;
else if (ll < lr)
ll = lr = ; if (rl == rr && (rl * ) > res)
res = * rl;
else if (rl > rr)
rl = rr = ; } return res;
}
private: };
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