@@ -26875,40 +26875,40 @@ func backtrack(left int, right int, track *string, res *[]string) {
2687526875```java
2687626876// by labuladong (java)
2687726877class Solution {
26878- public:
26879- vector<string > generateParenthesis(int n) {
26880- if (n == 0) return {} ;
26878+ 26879+ public List<String > generateParenthesis(int n) {
26880+ if (n == 0) return new ArrayList<>() ;
2688126881 // 记录所有合法的括号组合
26882- vector<string > res;
26882+ List<String > res = new ArrayList<>() ;
2688326883 // 回溯过程中的路径
26884- string track;
26884+ StringBuilder track = new StringBuilder() ;
2688526885 // 可用的左括号和右括号数量初始化为 n
2688626886 backtrack(n, n, track, res);
2688726887 return res;
2688826888 }
2688926889
2689026890 // 可用的左括号数量为 left 个,可用的右括号数量为 rgiht 个
2689126891 void backtrack(int left, int right,
26892- string& track, vector<string>& res) {
26892+ StringBuilder track, List<String> res) {
2689326893 // 若左括号剩下的多,说明不合法
2689426894 if (right < left) return;
2689526895 // 数量小于 0 肯定是不合法的
2689626896 if (left < 0 || right < 0) return;
2689726897 // 当所有括号都恰好用完时,得到一个合法的括号组合
2689826898 if (left == 0 && right == 0) {
26899- res.push_back (track);
26899+ res.add (track.toString() );
2690026900 return;
2690126901 }
2690226902
2690326903 // 尝试放一个左括号
26904- track.push_back ('('); // 选择
26904+ track.append ('('); // 选择
2690526905 backtrack(left - 1, right, track, res);
26906- track.pop_back( ); // 撤消选择
26906+ track.deleteCharAt(track.length() - 1 ); // 撤消选择
2690726907
2690826908 // 尝试放一个右括号
26909- track.push_back (')'); // 选择
26909+ track.append (')'); // 选择
2691026910 backtrack(left, right - 1, track, res);
26911- track.pop_back( ); // 撤消选择
26911+ track.deleteCharAt(track.length() - 1 ); // 撤消选择
2691226912 }
2691326913}
2691426914```
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