括号匹配问题

时间:2022-08-19 18:53:06

问题描述

Description

We give the following inductive definition of a “regular brackets” sequence:

  • the empty sequence is a regular brackets sequence,
  • if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
  • if a andb are regular brackets sequences, thenab is a regular brackets sequence.
  • no other sequence is a regular brackets sequence

描述

给你一个字符串,里面只包含"(",")","[","]"四种符号,
如果:
[]是匹配的
([])[]是匹配的
((]是不匹配的
([)]是不匹配的

For instance,all of the following character sequences are regular brackets sequences:

(), [], (()), ()[], ()[()]

while thefollowing character sequences are not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a1a2 … an,your goal is to find the length of the longest regular brackets sequence that is a subsequence ofs. That is, you wish to find the largestmsuch that for indicesi1,i2, …,imwhere 1 ≤i1 <i2 < … <imn,ai1ai2 …aim is a regular bracketssequence.

Given the initial sequence ([([]])], the longest regularbrackets subsequence is[([])].

Input

The input test file will contain multiple test cases. Each input test case consists of asingle line containing only the characters(,),[, and]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

Output

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

Sample Input

((()))
()()()
([]])
)[)(
([][][)
end

Sample Output

6
6
4
0
6


步骤分析

见代码实现步骤


代码实现

public class Brackets {
	private static String str = "([][][)" ;
	public static void main(String[] args) {
		//1、计算数据长度
		int slen = str.length();
		//2、生成动态规划数组dp
		int[][] dp = new int[slen][slen];
		//3、初始化dp  
		/* dp[i][j]:s[i]到s[j]之间包含的正确括号数
		 * i >= j  =>  dp[i][j] = 0;
		 * 因为数组初始化为0,这道题不用再初始化
		 */
		//4、动态规划dp
		/* 先算长度短的,再算长度长的
		 * s[i]匹配s[j]则 dp[i][j] = dp[i+1][j-1] + 2;
		 * 否则dp[i][j] = dp[i+1][j-1]
		 * 考虑到 ([])[] 这种情况,dp{[])[} = 2; dp{([])[]} 不等于 4而是等于dp{([])}+dp{[]} 
		 * 所以需要再  dp[i][len]内部进行依次内更新k=i:j-1 dp[i][j] = max{dp[i][k]+dp[k+1][j]};
		 */
		for (int len = 2; len <= slen; len++) 
			for (int i = 0; i + len - 1 < slen; i++) {
				int j = i + len - 1;
				if(match(str.charAt(i),str.charAt(j)))
					dp[i][j] = dp[i+1][j-1] + 2;
				for (int k = i; k < j; k++) 
					if(dp[i][j] < dp[i][k]+dp[k+1][j])
						dp[i][j] = dp[i][k]+dp[k+1][j];
			}
		System.out.println(dp[0][slen-1]);
//		//3、初始化dp  (以下的思路是有错误的)
//		/* dp[i][j]:s[i]到s[j]之间包含的正确括号数
//		 * i >= j  =>  dp[i][j] = 0;
//		 * 因为数组初始化为0,这道题不用再初始化
//		 */
//		System.out.println();
		
//		//4、动态规划dp
//		/* 
//		 * s[i]匹配s[j]则 dp[i][j] = dp[i+1][j-1] + 2;
//		 * 否则 dp[i][j] = dp[i+1][j-1]
//		 * 考虑到 ([])[] 这种情况,dp{[])[} = 2; dp{([])[]} 不等于 4而是等于dp{([])}+dp{[]} 
//		 * 所以需要再  dp[i][j]内部进行依次内更新 k=j:-1:i+1 dp[i][j] = max{dp[i][k]+dp[k+1][j]};
//		 */
//		for (int i = 0; i < len; i++) 
//			for (int j = i+1; j < len; j++) {
//				//dp更新
//				if(match(str.charAt(i),str.charAt(j)))
//					dp[i][j] = dp[i+1][j-1] + 2;
//				//dp[i][j]内更新
//				for (int k = i+1; k < j; k++) 
//					if(dp[i][j] < dp[i][k]+dp[k+1][j])
//						dp[i][j] = dp[i][k]+dp[k+1][j];
//			}
//		
//		
//		System.out.println(dp[0][len-1]);
	
	}
	
	
	private static boolean match(char l, char r) {
		return l=='(' && r==')'  ||  l=='[' && r==']';  //||  l=='{' && r=='}';
	} 

}

P.S.

需要至少添加多少个括号才能使这些括号匹配起来。

括号总数-最大匹配括号数。