Generate Parentheses

[解题思路]
典型的递归。一步步构造字符串。当左括号出现次数<n时，就可以放置新的左括号。当右括号出现次数小于左括号出现次数时，就可以放置新的右括号。

public class Solution {
    public ArrayList<String> generateParenthesis(int n) {
        ArrayList<String> result = new ArrayList<String>();
        StringBuilder builder = new StringBuilder();
        generate(result, builder, n, n);
        return result;
    }
    
    public void generate(ArrayList<String> result, StringBuilder builder, int start, int end){
        if(start == 0 && end == 0){
            result.add(builder.toString());
            return;
        }
        
        if(start > 0){
            builder.append('(');
            generate(result, builder, start-1, end);
            builder.deleteCharAt(builder.length()-1);
        }
        
        if(start < end){
            builder.append(')');
            generate(result, builder, start, end -1);
            builder.deleteCharAt(builder.length()-1);
        }
    }
}

注意：

builder.deleteCharAt(builder.length()-1); 
eg : n=2 时 builder 先是 “(( ))” ,添加到 result ArrayList中去
然后 delete 成为 “(”, 此时 end > start, 然后 builder 成为 “()” 在成为 “（）（）”