47. Permutations II**

47. Permutations II**

https://leetcode.com/problems/permutations-ii/

题目描述

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

C++ 实现 1

此题是 46. Permutations** 的进阶版本. 参考 46. Permutations** 的 C++ 实现 3 来实现此题. 注意在 dfs 函数中 nums 的类型 vector<int> 而不是 vector<int>& 引用类型, 这样在每次调用 nums 时都会发生一次拷贝. 倘若采用引用, if (i != start && nums[i] == nums[start]) continue; 的预期作用可能无法实现. 因为 nums

class Solution {
private:
    void dfs(vector<int> nums, int start, vector<vector<int>> &res) {
        if (start == nums.size()) {
            res.push_back(nums);
            return;
        }
        for (int i = start; i < nums.size(); ++ i) {
            if (i != start &&  nums[i] == nums[start]) continue;
            std::swap(nums[i], nums[start]);
            dfs(nums, start + 1, res);
        }
    }
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        std::sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        dfs(nums, 0, res);
        return res;
    }
};

错误代码

来看看一个错误示例, 分析错误的过程, 比如 [0, 0, 1, 1, 9] 这个例子:

class Solution {
private:
    void dfs(vector<int> &nums, int start, vector<vector<int>> &res) {
        if (start == nums.size()) {
            res.push_back(nums);
            return;
        }
        for (int i = start; i < nums.size(); ++ i) {
            if (i != start &&  nums[i] == nums[start]) continue;
            std::swap(nums[i], nums[start]);
            dfs(nums, start + 1, res);
            std::swap(nums[i], nums[start]);
        }
    }
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        std::sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        dfs(nums, 0, res);
        return res;
    }
};

最后的输出结果为:

0 0 1 1 9
0 0 1 9 1
0 0 9 1 1
0 1 0 1 9
0 1 0 9 1
0 1 1 0 9 *
0 1 1 9 0
0 1 9 1 0
0 1 9 0 1
0 1 1 0 9 *
0 1 1 9 0
....

上面带 * 的两行数据是重复的.

C++ 实现 2

来自 Accepted backtracking C++ solution by using map (28ms), 使用一个哈希表来消除重复元素.

class Solution {
public:
vector<vector<int> > permuteUnique(vector<int> &num) {
    vector<vector<int>> v;
    vector<int> r;
    map<int, int> map;
    for (int i : num)
    {
        if (map.find(i) == map.end()) map[i] = 0;
        map[i]++;
    }
    permuteUnique(v, r, map, num.size());
    return v;
}

void permuteUnique(vector<vector<int>> &v, vector<int> &r, map<int, int> &map, int n)
{
    if (n <= 0)
    {
        v.push_back(r);
        return;
    }
    for (auto &p : map)
    {
        if (p.second <= 0) continue;
        p.second--;
        r.push_back(p.first);
        permuteUnique(v, r, map, n - 1);
        r.pop_back();
        p.second++;
    }
}
};