LeetCode Day35 Permutation

在上一个答案的基础上，每个位置插入新的元素，构成新的排列

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        if (num.empty()) return vector<vector<int> >(1, vector<int>());
        vector<vector<int> > res;
        int first = num[0];
        num.erase(num.begin());
        vector<vector<int> > words = permute(num);
        for (auto &a : words) {
            for (int i = 0; i <= a.size(); ++i) {
                a.insert(a.begin() + i, first);
                res.push_back(a);
                a.erase(a.begin() + i);
            }
        }   
        return res;
    }
};

回溯。用递归DFS来求解。这里我们需要用到一个visited数组来标记某个数字是否访问过，然后在DFS递归函数每次递归循环从头开始：

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        vector<vector<int> > res;
        vector<int> out;
        vector<int> visited(num.size(), 0);
        permuteDFS(num, 0, visited, out, res);
        return res;
    }
    void permuteDFS(vector<int> &num, int level, vector<int> &visited, vector<int> &out, vector<vector<int> > &res) {
        if (level == num.size()) res.push_back(out);
        else {
            for (int i = 0; i < num.size(); ++i) {
                if (visited[i] == 0) {
                    visited[i] = 1;//标注为一作为头元素，下一步递归从剩下的不为一的值再循环
                    out.push_back(num[i]);
                    permuteDFS(num, level + 1, visited, out, res);
                    out.pop_back();
                    visited[i] = 0;
                }
            }
        }
    }
};

另一种递归写法，每次交换第一个元素

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        vector<vector<int> > res;
        permuteDFS(num, 0, res);
        return res;
    }
    void permuteDFS(vector<int> &num, int start, vector<vector<int> > &res) {
        if (start == num.size()) res.push_back(num);
        for (int i = start; i < num.size(); ++i) {
            swap(num[start], num[i]);//每次做头元素的值改变
            permuteDFS(num, start + 1, res);//交换后的序列除头元素之外的值递归
            swap(num[start], num[i]);//再返回原样
        }
    }
};