[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[n];
		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：

http://blog.csdn.net/linhuanmars/article/details/19873463

http://blog.csdn.net/fightforyourdream/article/details/14159435