46, 47 Permutations I, II-优快云博客

本文链接：https://blog.youkuaiyun.com/djgjmtdmjt/article/details/53954211

本文介绍了一种生成给定数字集合所有可能排列的算法，并针对含有重复数字的情况提出了优化方案，确保生成的排列结果唯一。

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Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:

[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

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]
]

代码：

class Solution {
public:
    /*vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> temp;
        vector<bool> used(nums.size(), false);
        helper(res, temp, nums.size(), nums, used);
        
        return res;
        
    }
    void helper(vector<vector<int>>& res, vector<int>& temp, int left, vector<int>& nums, vector<bool>& used){
        
        if(left == 0){
            res.push_back(temp);
            return;
        }
        for(int i = 0; i < nums.size(); i++){
            if(!used[i]){
                temp.push_back(nums[i]);
                used[i] = true;
                helper(res, temp, left - 1, nums, used);
                temp.pop_back();
                used[i] = false;
            }
        }
    }*/
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> res;
        helper(res, 0, nums);
        return res;
    }
    void helper(vector<vector<int>>& res, int begin, vector<int>& nums){
        if(begin >= nums.size()){
            res.push_back(nums);
            return;
        }
        for(int i = begin; i < nums.size(); i++){
            swap(nums[i], nums[begin]);
            helper(res, begin + 1, nums);
            swap(nums[i], nums[begin]);
        }
    }
};

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