[LeetCode]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.

［stack解法］要点：利用stack来记录‘（’的位置。

利用stack来存储最近未匹配‘（’的位置，如果遇到’）‘就出栈，如果栈为零说明前面都是匹配的，动态更新边界点，匹配完后记录长度，找到最大值。

空间O(N)，时间O(N)

class Solution {public:    int longestValidParentheses(string s) {        stack<int> stk; //no matching '('        int maxlen = 0;        int last = -1; //last ')'        for(int i=0; i<s.length(); ++i){            if(s[i]=='(')                stk.push(i);             else {                if(stk.empty()){                    last = i; //previous string match                }                else {                    stk.pop();                    if(stk.empty()){                        maxlen = max(maxlen, i-last); //start at ')' next                    }                    else                         maxlen = max(maxlen, i-stk.top()); //start at not match'(' next                }            }        }        return maxlen;    }};

［Dp解法］从字符串最大处开始往前扫。记录第一个未match的点，就是i+1+f(i+1)

如果未match的点和现在的点匹配，说明f(i) = f(i+1)+2;

同时还要注意一种情况，未和现在点match的后面可能还有解。比如((()))(),这种类型，此时f(i) += f(match+1)才是最后解。

class Solution {public:    int longestValidParentheses(string s) {        vector<int> Dp(s.size(),0);        int maxlen;        if(s.size()<2)            return 0;        for(int i=s.size()-2; i>=0; --i){            int match = Dp[i+1]+i+1; //the last match ')'            if(s[i]=='(' && match<s.size()&&s[match]==')'){ //match<s.size() not the last                Dp[i] = Dp[i+1]+2;                if(match+1<s.size()){ //((()))(....)                    Dp[i] += Dp[match+1];                }            }            maxlen = max(Dp[i],maxlen);        }        return maxlen;    }};