JZ27 字符串的排列

本文最后更新于：2022年4月9日 中午

image-20211008103531881

Solution

  • 回溯法
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
class Solution {
public:
    vector<string> Permutation(string s) {
        int n = s.size();
        if (n == 0) return res;
        sort(s.begin(), s.end());

        string record;
        used = vector<bool>(n, false);
        backtrack(s, record);
        return res;
    }

private:
    vector<bool> used;
    vector<string> res;

    void backtrack(string& s, string& record) {
        if (record.size() == s.size()) {
            res.push_back(record);
            return;
        }

        for (int i = 0; i < s.size(); ++i) {
            if (!used[i]) {
                if (i != 0 && s[i] == s[i-1] && used[i-1] == false)
                    continue;
                used[i] = true;
                record.push_back(s[i]);
                backtrack(s, record);
                record.pop_back();
                used[i] = false;
            }
        }
    }
};
  • 借助库函数
1
2
3
4
5
6
7
8
9
10
11
12
13
class Solution {
public:
    // 借助系统自带函数 next_permutation
    vector<string> Permutation(string str) {
        vector<string> res;
        if (str.size() == 0) return res;
        sort(str.begin(), str.end());
        do {
            res.push_back(str);
        } while (next_permutation(str.begin(), str.end()));
        return res;
    }
};