47. 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]
]

不会 学习：https://leetcode.com/problems/permutations-ii/discuss/

class Solution {
public:
    vector<vector<int> > permuteUnique(vector<int> &nums) {
        // res.clear();
        sort(nums.begin(), nums.end());        //排序
        res.push_back(nums);      //第一个排列是自身
        int j;
        int i = nums.size()-1;
        while (1){
            for (i=nums.size()-1; i>0; i--){
                if (nums[i-1]< nums[i]){    //从后往前找到逆序结束的地方[1,1,2], [1,2,1]. no：[2,1,1]. 
                     break;
                }
            }
            if(i == 0){        //说明全部排列都没了[2,1,1]
                break;
            }

            for (j=nums.size()-1; j>i-1; j--){   
                if (nums[j]>nums[i-1]){       //[1,1,2],  i=1->2  j=2->1
                    break;
                }
            }                    
            swap(nums[i-1], nums[j]);         //[1,1,2]->[1,2,1]
            reverse(nums, i, nums.size()-1);
            res.push_back(nums);
        }
        return res;
    }
    void reverse(vector<int> &nums, int s, int e){         
        while (s<e){                        //使得排列13245比13542先加入res  13542->13245 
            swap(nums[s++], nums[e--]);
        }
    }
    
    vector<vector<int> > res;
};

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;  //避免重复
            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;
    }
};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;  //避免重复
            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;
    }
};