[LeetCode] Permutations II

 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]
]

给定一个包含重复元素的数组,找出其所有全排列中不重复的那些组合。

思路:使用backtracking算法。

helper函数用于寻找下一个元素,swap用于构造全排列。

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> res;
        int idx = 0;
        helper(res, nums, idx);
        return res;
    }
    
    void helper(vector<vector<int>>& res, vector<int>& nums, int idx) {
        if (idx >= nums.size()) {
            if (find(res.begin(), res.end(), nums) == res.end()) {
                res.push_back(nums);
            }
        }
        else {
            for (int i = idx; i < nums.size(); i++) {
                swap(nums[i], nums[idx]);
                helper(res, nums, idx + 1);
                swap(nums[i], nums[idx]);
            }
        }
    }
};
// 469 ms
原文地址:https://www.cnblogs.com/immjc/p/8359073.html