LeetCode_22 Generate Parentheses-优快云博客

本文链接：https://blog.youkuaiyun.com/sbicode/article/details/50598205

本文介绍了一种递归算法来生成所有可能的有效括号组合。针对给定的n对括号，通过控制左括号和右括号的数量，确保生成的括号序列是合法的。该算法使用了回溯思想，并提供了详细的Java实现代码。

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Link to original problem: 这里写链接内容
Very typical backtracking problem, it can be simulated to give all the possible orders of n 0s and n 1s list. But we need to take care about that at every index of each output, ‘)’ can not over number than the left ‘(‘.

So we add canLeft and canRight to control the whole process. Initially canLeft = n and canRight = 0. When we add a ‘(‘, canLeft- - and canRight++, when we add a ‘)’, only canRight- -.

Related problem is 17 LetterCombinations of a Phone Number 这里写链接内容

Here is the description of 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:
“((()))”, “(()())”, “(())()”, “()(())”, “()()()”

Here is the recursive solution:

public class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<String>();
        StringBuffer cur = new StringBuffer();
        helper(res, cur, 0, n);
        return res;
    }
    private void helper(List<String> res, StringBuffer cur, int canRight, int canLeft){
        if(canRight == 0 && canLeft == 0)
            res.add(cur.toString());
        else{
            if(canLeft == 0){
                cur.append(')');
                helper(res, cur, canRight-1, canLeft);
                cur.deleteCharAt(cur.length()-1);
            }
            else{
                cur.append('(');
                helper(res, cur, canRight+1, canLeft-1);
                cur.deleteCharAt(cur.length()-1);
                if(canRight != 0){
                    cur.append(')');
                    helper(res, cur, canRight-1, canLeft);
                    cur.deleteCharAt(cur.length()-1);
                }
            }
        }
    }
}