LeetCode | 47. Permutations II-优快云博客

本文介绍了解决含有重复元素集合的所有唯一排列的方法。提供了两种递归算法实现：一种使用set进行结果去重；另一种通过排序并跳过重复元素来避免重复排列。

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

For example,
[1,1,2] have the following unique permutations:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

思路一：递归，用 set 去重

//63 ms
class Solution {
public:
    set <vector<int> > result;

    // permute num[begin..end]
    // invariant: num[0..begin-1] have been fixed/permuted
    void permuteRecursive(vector<int> &num, int begin, set<vector<int> > &result)
    {
        if (begin >= num.size()) {
            // one permutation instance
            result.insert(num);
            return;
        }

        for (int i = begin; i < num.size(); i++) {
            swap(num[begin], num[i]);
            permuteRecursive(num, begin + 1, result);
            // reset
            swap(num[begin], num[i]);
        }
    }

    vector<vector<int> > permuteUnique(vector<int> &num)
    {
        vector<vector<int> > res;
        permuteRecursive(num, 0, result);
        set<vector<int> >::iterator it = result.begin();
        while(it != result.end())
        {
            res.push_back(*it);
            it++;
        }
        return res;
    }
};

思路二：递归，不相同的元素可以交换

//22 ms
class Solution {
public:
    void recursion(vector<int> num, int i, int j, vector<vector<int> > &res) {
        if (i == j-1) {
            res.push_back(num);
            return;
        }
        for (int k = i; k < j; k++) {
            if (i != k && num[i] == num[k]) continue;
            swap(num[i], num[k]);
            recursion(num, i+1, j, res);
        }
    }
    vector<vector<int> > permuteUnique(vector<int> &num) {
        sort(num.begin(), num.end());
        vector<vector<int> >res;
        recursion(num, 0, num.size(), res);
        return res;
    }
};