2014
11-18

# LeetCode-Longest Valid Parentheses[栈]

### 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.

// LeetCode, Longest Valid Parenthese
// 使用栈，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
int longestValidParentheses(string s) {
int max_len = 0, last = -1; // the position of the last ')'
stack<int> lefts;  // keep track of the positions of non-matching '('s

for (int i = 0; i < s.size(); ++i) {
if (s[i] =='(') {
lefts.push(i);
} else {
if (lefts.empty()) {
// no matching left
last = i;
} else {
// find a matching pair
lefts.pop();
if (lefts.empty()) {
max_len = max(max_len, i-last);
} else {
max_len = max(max_len, i-lefts.top());
}
}
}
}
return max_len;
}
};


// LeetCode, Longest Valid Parenthese
// 时间复杂度O(n)，空间复杂度O(n)
// @author 一只杰森(http://weibo.com/wjson)
class Solution {
public:
int longestValidParentheses(string s) {
vector<int> f(s.size(), 0);
int ret = 0;
for (int i = s.size() - 2; i >= 0; --i) {
int match = i + f[i + 1] + 1;
// case: "((...))"
if (s[i] == '(' && match < s.size() && s[match] == ')') {
f[i] = f[i + 1] + 2;
// if a valid sequence exist afterwards "((...))()"
if (match + 1 < s.size()) f[i] += f[match + 1];
}
ret = max(ret, f[i]);
}
return ret;
}
};


// LeetCode, Longest Valid Parenthese
// 两遍扫描，时间复杂度O(n)，空间复杂度O(1)
// @author 曹鹏(http://weibo.com/cpcs)
class Solution {
public:
int longestValidParentheses(string s) {
int answer = 0, depth = 0, start = -1;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(') {
++depth;
} else {
--depth;
if (depth < 0) {
start = i;
depth = 0;
} else if (depth == 0) {
}
}
}

depth = 0;
start = s.size();
for (int i = s.size() - 1; i >= 0; --i) {
if (s[i] == ')') {
++depth;
} else {
--depth;
if (depth < 0) {
start = i;
depth = 0;
} else if (depth == 0) {
}
}
}
}
};


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应改为L（X [0 .. M-1]，Y [0 .. N-1]）= 1 + L（X [0 .. M-2]，Y [0 .. N-2]）

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