題目

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:

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

代碼

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

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* 本文來自博客 ?“李博Garvin“

* 轉(zhuǎn)載請(qǐng)標(biāo)明出處:http://blog.csdn.net/buptgshengod

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