Longest Valid parentheses (length of the longest valid match bracket substring)

Transferred from: http://blog.csdn.net/cfc1243570631/article/details/9304525

Title Description:

Given A string containing just the characters ' (' and ') ', find the length of the longest valid (well-formed) parentheses Su Bstring.

For "(()", the longest valid parentheses substring are "()", which has length = 2.

Another example is ") () ()", where the longest valid parentheses substring are "() ()", which has length = 4.

Translation:

Given a string containing ' (' and ') ', find out the length of the longest valid brace matching substring.

Solution:

This problem can be solved with one-dimensional dynamic programming in inverse way. Suppose the input parenthesis expression is string s, maintaining a one-dimensional array of length s.length dp[], and array elements initialized to 0. Dp[i] represents the longest valid matching brace substring length from s[i] to S[s.length-1] that contains s[i]. The following relationship exists: dp[s.length-1] = 0; I from n-2, 0 reverse dp[], and record its maximum value. If s[i] = = ' (', then the value of Dp[i] is calculated from I to s.length-1 in S. This calculation is divided into two steps, via Dp[i + 1] (note dp[i + 1] has been solved in the previous step): Looking for a valid brace matching substring length starting from i + 1 in S, that is, Dp[i + 1], skipping this valid brace substring, viewing the next character, with the subscript j = i + 1 + dp[i + 1]. If J does not cross over and s[j] = = ') ', then S[i ... j] is a valid brace match, dp[i] =dp[i + 1] + 2. After a valid match length of s[i ... j] is obtained, if J + 1 does not cross the bounds, then the value of Dp[i] is added to the longest valid match starting with J + 1, i.e. Dp[j + 1].

[CPP]  View plain copy #include  <stdio.h>   #include  <string.h>   #define  N 65536   int longestvalidparentheses (const char *s)    {        int i,j,n;       int dp[N];       int max=0;       n=strlen (s);        for (i=0;i<n;i++)            dp[i]=0;        for (i=n-2;i>=0;i--)        {           if (s[i]== ' (')             {               j=i+1+dp[i+1] ;               if (j<n &&  s[j]== ') ')                {                    dp[i]=dp[i+1]+2;                    if (j+1<n)                         dp[i]+=dp[j+1];                }           }            if (Max<=dp[i])                 max=dp[i];       }        return max;  }   int main ()    {       &NBSP;&NBSP;&NBSP;     const char *s= ")) () ()";       printf ("% D\n ", longestvalidparentheses (s));       return 0;  }