LeetCode: Generate Parentheses

第一次memory exceed了，想复杂了，看了答案才知道只要函数系数多点就行了。。

class Solution {
public:
    void dfs(int dep, int leftnum, int n, vector<string> &ret, int leftnumtotal, string tmp) {
        if (leftnumtotal > n) return;
        if (dep == 2*n) {
            ret.push_back(tmp);
            return;
        }
        dfs(dep+1, leftnum+1, n, ret, leftnumtotal+1, tmp+'(');
        if (leftnum > 0)
            dfs(dep+1, leftnum-1, n, ret, leftnumtotal, tmp+')');
    }
    vector<string> generateParenthesis(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> ret;
        string tmp = "";
        if (!n) return ret;
        dfs(0, 0, n, ret, 0, tmp);
        return ret;
    }
};

 也可以用下面这段bfs的算法

struct par {
    string s;
    int total;
    int left;
    par() : s(""), total(0), left(0) { }
    par(string a, int b, int c) : s(a), total(b), left(c) { }
};
    
class Solution {
public:
    vector<string> generateParenthesis(int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if (n == 0) return vector<string>(0);
        queue<par> que;
        vector<string> res;
        que.push(par("(", 1, 1));
        while (!que.empty()) {
            par front = que.front();
            que.pop();
            if (front.total == n && front.left == 0) res.push_back(front.s);
            else {
                if (front.total < n) {
                    que.push(par(front.s+"(", front.total+1, front.left+1));
                    if (front.left > 0) que.push(par(front.s+")", front.total, front.left-1));
                }
                else que.push(par(front.s+")", front.total, front.left-1));
            }
        }
        return res;
    }
};

 C#

public class par {
    public string s;
    public int total;
    public int left;
    public par(string a, int b, int c) { s = a; total = b; left = c; }
}
public class Solution {
    public List<string> GenerateParenthesis(int n) {
        List<string> ans = new List<string>();
        if (n == 0) return ans;
        Queue<par> que = new Queue<par>();
        que.Enqueue(new par("(", 1, 1));
        while (que.Count != 0) {
            par peek = que.Peek();
            que.Dequeue();
            if (peek.total == n && peek.left == 0) ans.Add(peek.s);
            else {
                if (peek.total < n) {
                    que.Enqueue(new par(peek.s+"(", peek.total+1, peek.left+1));
                    if (peek.left > 0) que.Enqueue(new par(peek.s+")", peek.total, peek.left-1));
                }
                else que.Enqueue(new par(peek.s+")", peek.total, peek.left-1));
            }
        }
        return ans;
    }
}