**Leetcode 32. Longest Valid Parentheses

主要两种思路，stack 以及 dp

我想出来的是stack的solution 自己的Dp没A掉，看了别人的Dp思路才懂。

1、最简单易懂的

struct Node {int idx;char c;Node(int idx, char c): idx(idx), c(c) {}};class Solution {public:int match(char l, char r) {if ( l == '(' && r == ')' ) {return 2;} else {return 0;}}int longestValidParentheses(string s) {int ans = 0;stack<Node> sta;for (int i = 0; i < s.size(); i++) {if ( sta.size() == 0 || !match( sta.top().c, s[i] ) ) {sta.push(Node(i, s[i]));} else {sta.pop();}}sta.push(Node(s.size(), '|'));int idxs[sta.size()];int len = sta.size();for ( int i = sta.size()-1; i >= 0; i-- ) {idxs[i] = sta.top().idx;sta.pop();}int last_v = 0;for ( int i = 0; i < len; i++ ) {ans = max( idxs[i] - last_v, ans );last_v = idxs[i] + 1;}return ans;}};

2、干净的stack代码

class Solution {public:    int longestValidParentheses(string s) {        stack <int> sta;        int ans = 0, l = 0;        for (int i = 0; i < s.size(); i++) {            if (s[i] == '(') {                sta.push(i);            } else {                if (sta.empty()) {                    l = i + 1;                } else {                    sta.pop();                    if (sta.empty()) {                        ans = max( ans, i - l + 1 );                    } else {                        ans = max( ans, i - sta.top() );                    }                }            }        }        return ans;    }};

3、dp

class Solution {public:    int longestValidParentheses(string s) {        int ans = 0;        if (s.size() < 2) return 0;        int longest[s.size()];                for (int i = 0; i < s.size(); i++) {            longest[i] = 0;            if (s[i] == ')' && i - 1 >= 0 && i - longest[i - 1 ] - 1 >= 0 && s[ i - longest[i - 1 ] - 1 ] == '(' ) {                longest[i] = longest[i-1] + 2 + ( (i-longest[i-1]-2 >= 0)?longest[i-longest[i-1]-2]:0 );                ans = max(ans, longest[i]);            }        }        return ans;    }};