[leetcode] Generate Parentheses

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"

https://oj.leetcode.com/problems/generate-parentheses/

思路1：自己模仿permutation之类的题目，一个一个往数据里面填，填的时候判断是否是valid的，每次都要统计括号数，麻烦。

思路2：改进思路1，也是递归实现，将已有的左右括号数作为参数递归下去，写起来很简洁。

思路1：

public class Solution {
public ArrayList<String> generateParenthesis(int n) {
ArrayList<String> result = new ArrayList<String>();
if (n <= 0)
return result;
n = n * 2;
char str[] = new char
;
generate(0, n, str, result);

return result;
}

private void generate(int cur, int n, char[] s, ArrayList<String> result) {
if (cur == n) {
result.add(new String(s));
} else {
int j;
int cntL = 0, cntR = 0;
for (j = 0; j < cur; j++) {
if (s[j] == '(')
cntL++;
else
cntR++;
}
if (cntL > cntR) {
s[cur] = ')';
generate(cur + 1, n, s, result);

if (cntL < n / 2) {
s[cur] = '(';
generate(cur + 1, n, s, result);
}
} else if (cntL == cntR) {
s[cur] = '(';
generate(cur + 1, n, s, result);
} else {
return;
}

}

}

public static void main(String[] args) {
System.out.println(new Solution().generateParenthesis(3));
}

}

思路2：

import java.util.ArrayList;

public class Solution {
public ArrayList<String> generateParenthesis(int n) {
ArrayList<String> res = new ArrayList<String>();
if (n <= 0)
return res;
StringBuilder sb = new StringBuilder();
generate(n, n, sb, res);
return res;
}

private void generate(int l, int r, StringBuilder sb, ArrayList<String> res) {
if (r < l)
return;
if (l == 0 && r == 0) {
res.add(sb.toString());
}
if (l > 0) {
sb.append("(");
generate(l - 1, r, sb, res);
sb.deleteCharAt(sb.length() - 1);
}
if (r > 0) {
sb.append(")");
generate(l, r - 1, sb, res);
sb.deleteCharAt(sb.length() - 1);
}
}

public static void main(String[] args) {
System.out.println(new Solution().generateParenthesis(3));
}

}

第二遍记录：将已经用过的( 和 ) 传递下去，根据数量判断是否要继续添加，最后注意终止条件的(和)的数目都是n。

public class Solution {
public List<String> generateParenthesis(int n) {
List<String> res = new ArrayList<String>();
if (n <= 0)
return res;
StringBuilder sb = new StringBuilder();
generate(0, 0, sb, res, n);
return res;
}

private void generate(int lUsed, int rUsed, StringBuilder sb, List<String> res, int n) {
if (lUsed < rUsed)
return;
if (lUsed == n && rUsed == n) {
res.add(sb.toString());
}
if (lUsed < n) {
sb.append("(");
generate(lUsed + 1, rUsed, sb, res, n);
sb.deleteCharAt(sb.length() - 1);
}

if (rUsed < n) {
sb.append(")");
generate(lUsed, rUsed + 1, sb, res, n);
sb.deleteCharAt(sb.length() - 1);
}

}

}

第三遍记录： 递推向下时，忘记判断 l>0 和 r>0了。 否则 l和r一直减一变成负数了。。

if (l > 0) {
sb.append("(");
generate(l - 1, r, sb, res);
sb.deleteCharAt(sb.length() - 1);
}

第四遍记录：

先填(，注意不能超过n个，再填)，注意不能超过n，也不能超过前面的(的数量。填满到2n保存结果。

import java.util.ArrayList;
import java.util.List;

public class Solution {
public List<String> generateParenthesis(int n) {
List<String> res = new ArrayList<String>();
StringBuilder sb = new StringBuilder();
if (n <= 0)
return res;
geneParen(0, 0, n, sb, res);

return res;
}

private void geneParen(int left, int right, int n, StringBuilder sb, List<String> res) {
if (sb.length() == 2 * n) {
res.add(sb.toString());
return;
}
if (left < n) {
sb.append('(');
geneParen(left + 1, right, n, sb, res);
sb.deleteCharAt(sb.length() - 1);
}

if (right < n && right < left) {
sb.append(')');
geneParen(left, right + 1, n, sb, res);
sb.deleteCharAt(sb.length() - 1);
}

}

public static void main(String[] args) {
System.out.println(new Solution().generateParenthesis(5));
}
}

参考：

/article/1378361.html

/article/1347010.html