DFS

public class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> results = new ArrayList<>();
        if (n == 0) {
            return results;
        }
        generate(0, 2 * n, "", results);
        return results;
    }
    private void generate(int level, int n, String curState, List<String> results) {
        //recursion terminstor 先写
        if (level >= n) {
            //方法一：此处判断是否合法
            if (isValid(curState)) {
                results.add(curState);
            }
            
            return;
        }
        generate(level + 1, n, curState + "(", results);
       
        generate(level + 1, n, curState + ")", results);
    }
    private boolean isValid(String s) {
        char[] c = s.toCharArray();
        Stack stack = new Stack();
        for (int i = 0; i < c.length; i++) {
            if(c[i] == ')' && !stack.isEmpty()) {
                stack.pop();
            } else if (c[i] == '(') {
                stack.push(c[i]);
            } else {
                return false;
            }
        }
        if (stack.isEmpty()) {
            return true;
        }
        return false;
    }
}

O(T)有问题

应该每步进行判断。

public class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> results = new ArrayList<>();
        if (n == 0) {
            return results;
        }
        gen(n, n, "", results);
        return results;
    }
    private void gen(int left, int right, String cur, List<String> results) {
        if (left == 0 && right == 0) {
            results.add(cur);
        }
        if (left > right) {
            return;
        }
        if (left > 0) {
            gen(left - 1, right, cur + '(', results);
        }
        if (right > 0 ) {
            gen(left, right - 1, cur + ')', results);
        }
    }
}

思路：先有左括号，再有右括号。