[Leetcode] 032. Longest Valid parentheses (hard) (C + +)

Index: [Leetcode] Leetcode indicator interpretation (C++/JAVA/PYTHON/SQL)
Github:https://github.com/illuz/leetcode

032. Longest Valid parentheses (hard) links：

Title: https://oj.leetcode.com/problems/longest-valid-parentheses/
Code (GitHub): Https://github.com/illuz/leetcode

Test Instructions：

Ask the length of the longest legal parenthesis string in a

Analysis：

1. (c + +) is done with a stack, assuming the match is out of the stack, and then the length is the cur - stack_top_pos previous position of the match. O (n) time, O (n) space.
2. (c + +) stack consumes too much space. can actually maintain () the length of the match. It is only ())) possible ((() to appear and the situation. So you have to sweep it over and over.

   O (n) time, O (1) space.

3. With a more complex DP to do, but the space can not solve the solution 2 so excellent. Just saw me a very long time ago a solution, with too much space for Orz. Now it is 1 or 2 good practice.

Code：

Solution 1: (c + +)

Class Solution {public:    int longestvalidparentheses (string s) {        stack<int> lefts;        int Max_len = 0, Match_pos =-1;    Position of first                                            //matching ' ('-1 for        (int i = 0; i < s.size (); ++i) {            if (s[i] = = ' (')                Lefts.pu SH (i);            else {                if (Lefts.empty ())  //No matching left                    match_pos = i;                else {              //match a left                    Lefts.pop ();                    if (Lefts.empty ())                        Max_len = max (Max_len, i-match_pos);                    else                        Max_len = max (Max_len, I-lefts.top ());        }}} return max_len;}    ;

Solution 2: (c + +)

Class Solution {Public:int longestvalidparentheses (string s) {int max_len = 0, depth = 0, start =-1;            Solve (() for (int i = 0; i < s.size (); ++i) {if (s[i] = = ' (') ++depth;                else {--depth;                if (depth = = 0) Max_len = max (Max_len, I-start);                    else if (Depth < 0) {start = i;                depth = 0;        }}}//Solve ())) depth = 0;        Start = S.size ();            for (int i = s.size (); I >= 0; i) {if (s[i] = = ') ') ++depth;                else {--depth;                if (depth = = 0) Max_len = max (Max_len, start-i);                    else if (Depth < 0) {start = i;                depth = 0;    }}} return Max_len; }};

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[Leetcode] 032. Longest Valid parentheses (hard) (C + +)