题解 | #有重复项数字的全排列#

#include <vector>
class Solution {
public:
    vector<vector<int> > permuteUnique(vector<int> &num) {
        sort(num.begin(), num.end());
        int n = num.size();
        vector<vector<int> > res;
        vector<int> path;
        vector<int> visited(n, 0);
        path.reserve(n);
        function<void()> dfs = [&]() {
            if (path.size() == n) {
                res.push_back(path);
                return;
            }
            for (int i = 0; i < n; ++i) {
                if (visited[i] || i > 0 && num[i - 1] == num[i] && !visited[i - 1]) {
                    continue;
                }
                visited[i] = 1;
                path.push_back(num[i]);
                dfs();
                path.pop_back();
                visited[i] = 0;
            }
        };
        dfs();
        return res;
    }
};

思路：dfs搜索，去重剪枝。