[LeetCode] Generate Parentheses

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:

["((()))","(()())","(())()","()(())","()()()"
]

括号匹配。使用递归的方法。

设定left、right表示左右括号的剩余数，

如果left > right表示匹配失衡，直接返回。

如果left < right表示已经存在一个左括号，需要匹配下一次左括号或者右括号，进行递归。

如果left == right == 0，表示括号匹配完毕且合法。把当前括号序列放入结果数组中。

参考代码如下：

class Solution {
public:vector<string> generateParenthesis(int n) {vector<string> res;generateParenthesis(n, n, "", res);return res;}void generateParenthesis(int left, int right, string str, vector<string>& res){if (left > right){return;}if (left == 0 && right == 0){res.push_back(str);   }else{if (left > 0){generateParenthesis(left-1, right, str+'(', res);}if (right > 0){generateParenthesis(left, right-1, str+')', res);}}}
};

另一种方法，提前剪枝，思路更清晰,left、right表示当前左右括号的数量

class Solution {
public:vector<string> generateParenthesis(int n) {vector<string> res;string out;dfs(n, 0, 0, out, res);return res;}void dfs(int n, int left, int right, string out, vector<string>& res){if (left < n){out.push_back('(');dfs(n, left+1, right, out, res);out.pop_back();}if (left > right){out.push_back(')');dfs(n, left, right+1, out, res);out.pop_back();}if (out.size() == n*2)res.push_back(out);}
};

转载于:https://www.cnblogs.com/immjc/p/9454370.html

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