给出 n 代表生成括号的对数,请你写出一个函数,使其能够生成所有可能的并且有效的括号组合。
例如,给出 n = 3,生成结果为:
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
Java代码:
public List<String> generateParenthesis(int n) {
List<String> ans = new ArrayList<String>();
backtrack(ans, "", 0, 0, n);
return ans;
}
public void backtrack(List<String> ans, String cur, int left, int right, int max){
if (cur.length() == max * 2) {//字符串长度达到最长时退出
ans.add(cur);
return;
}
if (left < max)
backtrack(ans, cur+"(", left+1, right, max);
if (right < left)
backtrack(ans, cur+")", left, right+1, max);
}
Python代码:
def generateParenthesis(self, N):
ans = []
def backtrack(S = '', left = 0, right = 0):
if len(S) == 2 * N:
ans.append(S)
return
if left < N:
backtrack(S+'(', left+1, right)
if right < left:
backtrack(S+')', left, right+1)
backtrack()
return ans
这篇博客对回溯法写的还蛮好的,大家可以参考一下https://blog.csdn.net/sinat_27908213/article/details/80599460