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:

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

代码

For 2, it should place one "()" and add another one insert it but none tail it,

'(' f(1) ')' f(0)

or add none insert it but tail it by another one,

'(' f(0) ')' f(1)

Thus for n, we can insert f(i) and tail f(j) and i+j=n-1,

'(' f(i) ')' f(j)

public List
     
       generateParenthesis(int n) {    List
      
        result = new ArrayList
       
        ();    if (n == 0) {        result.add("");    } else {        for (int i = n - 1; i >= 0; i--) {            List
        
          insertSub = generateParenthesis(i);            List
         
           tailSub = generateParenthesis(n - 1 - i);            for (String insert : insertSub) {                for (String tail : tailSub) {                    result.add("(" + insert + ")" + tail);                }            }        }    }    return result;}