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

Index: [Leetcode] leetcode key index (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, and if the match is out of the stack, 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, in fact, can maintain () the length of the match, but may appear ())) and ((() the situation, so to sweep back and forth. 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 long time ago a solution, with too much space Orz. Now it's 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; }};

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