NYOJ 15- brace matching (b)

15 matching parentheses (b)

Memory limit: 64MB  Time limit: 1000ms  special sentence: No
by number: 91  Submissions: 276  Difficulty: 6

Subject description:

Give you a string, which contains only "(", ")" "[", "]" four kinds of symbols, how many would you need to add at least these brackets parentheses match up.
Such as:
[] is matched
([]) [] is matched
((] is not matched
([)] is mismatched

Enter a description:

The first line enter a positive integer N, the number of test data (N <= 10) sets 
each have only one row of test data, a string S, S containing only four characters mentioned above, no length S More than 100

Output Description:

For each set of test data is output a positive integer indicating the minimum number of brackets to be added. Output row for each test

Sample input:

copy

4
[]
([])[]
((]
([)]

Sample output:

0 
0 
. 3 
2 

Reference LeetCode32 longest valid parentheses solution.

LeetCode32 longest valid parentheses using methods remove all perfect string matching substring After the string is assumed treatment stringBuffer. On the basis of this string will be much simpler to do. 
For perfect no substring string s, using d [i] [j] represents the substring [i, j) require a minimum number of brackets added. Note that strings like need another determination, such as (]) 

state transition equation is:

 

 

 1 import java.util.Scanner;
 2 
 3 public class Main{
 4     static class Solution {
 5         boolean match(char a, char b) {
 6             if (a == '(' && b == ')' || a == '[' && b == ']')
 7                 return true;
 8             return false;
 9         }
10 
11         public int solve(String s) {
12             if (s.length()<=1) return s.length();
13             int[][] d = new int[s.length()+1][s.length()+1];
14 
15             for (int i = 0; i < s.length(); i++) {
16                 d[i][i] = 0;
17                 d[i][i + 1] = 1;
18             }
19 
20             for (int step = 2; step <= s.length(); step++) {
21                 for (int i = 0; i < s.length() && i+step<s.length()+1; i++) {
22                     if (match(s.charAt(i), s.charAt(i + step - 1))) {
23                         d[i][i+step]=d[i+1][i+step-1];
24                     }else {
25                         int min = 65535;
26                         for (int k = i + 1; k < i + step; k++) {
27                             min = Math.min(d[i][k] + d[k][i + step], min);
28                         }
29                         d[i][i + step] = min;
30                     }
31                 }
32             }
33             return d[0][s.length()];
34         }
 35      }
 36  
37 [      public  static the StringBuffer deSubPerfect (String S) {
 38 is          int [] D = new new  int [s.length ()];
 39          int max = 0 ;
 40  
41 is          for ( int I =. 1; I <s.length (); i ++ ) {
 42 is              // find the most perfect i matches the end position 
43 is              int len = 0 ;
 44 is              IF (. 1-i-D [. 1-i]> = 0 && (s.charAt (i) == ')' && '(' == s.charAt (i - 1 - d [i - 1]) || s.charAt (i) == ']' && '[' == s.charAt (i-1 -d [-I. 1]))) { // (()) 
45                 len=d[i-1]+2;
46             }else if (!(s.charAt(i)==')' && s.charAt(i-1)=='(') || !(s.charAt(i)==']' && '['==s.charAt(i-1-d[i-1]))){// ())
47                 len=0;
48             }else {// (() || ()
49                 len = 2;
50             }
51 
52             if (i-len>=0 && len!=0){
53                 d[i]=d[i-len]+len;
                 d [i] =55{
else             }54len;
56             }
57             max = (max < d[i]) ? d[i] : max;
58         }
59         StringBuffer s1 = new StringBuffer();
60         for (int i = s.length()-1; i >= 0; i--) {
61             i-=d[i];
62             if (i>=0)
63                 s1.append(s.charAt(i));
64         }
65         StringBuffer reverse = s1.reverse();
66         return reverse;
67     }
68 
69     public static void main(String[] args) {
70         String s = new String();
71         Scanner sc = new Scanner(System.in);
72         Solution solution = new Solution();
73         int n = Integer.parseInt(sc.nextLine());
74         for (int i = 0; i < n; i++) {
75             s=sc.nextLine();
76             StringBuffer stringBuffer = deSubPerfect(s);
77 
78             int solve = solution.solve(stringBuffer.toString());
79             System.out.println(solve);
80         }
81     }
82 }
83 //()(]()
84 //(())(())