The set [1,2,3,…,_n_] contains a total of n! unique permutations.
By listing and labeling all of the permutations in order, We get the following sequence (ie, for n = 3):
1. `"123"` 2. `"132"` 3. `"213"` 4. `"231"` 5. `"312"` 6. `"321"` Given n and k, return the _k_th permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
思路：还是根据leetcode 31 解题报告 中的算法搞一下就好了。。
Given a collection of numbers that might contain duplicates, return all possible unique permutations.__
思路：和leet code 46 类似，最后用set去个重即可。。
代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月13日 星期四 15时00分48秒 4File Name :47.cpp 5************************************************ */ 6class Solution { 7 8public: 9 void solve( vector<int>&nums) 10 { 11 int n = nums.size(); 12 if (n==0) return; 13 int k = -1; 14 for ( int i = n-2 ; i >= 0 ; i--) 15 { 16 if (nums[i]<nums[i+1]) 17 { 18 k = i; 19 break; 20 } 21 } 22 if (k==-1) 23 { 24 reverse(nums.begin(),nums.end()); 25 return; 26 } 27 int l = -1; 28 for ( int i = n-1 ; i >k ; i--) 29 { 30 if (nums[k]<nums[i]) 31 { 32 l = i; 33 break; 34 } 35 } 36 swap(nums[l],nums[k]); 37 reverse(nums.begin()+k+1,nums.end()); 38 } 39 40 void pr (vector<int> &nums) 41 { 42 int siz = nums.size(); 43 for ( int i = 0 ; i < siz; i++) 44 printf("%d%c",nums[i],i==siz-1?'\n':' '); 45 } 46 vector<vector<int>> permuteUnique(vector<int>& nums) { 47 set<vector<int> >se; 48 vector<vector<int> >res; 49 int n = nums.size(); 50 int total = 1 ; 51 for ( int i = 2 ; i <= n ; i++) total*=i; 52 53 for ( int i = 1 ; i <= total ; i++) 54 { 55 se.insert(nums); 56// pr(nums); 57 solve(nums); 58 } 59 for ( auto &it :se) 60 { 61 res.push_back(it); 62 } 63 return res; 64 } 65 66};
Given a collection of distinct numbers, return all possible permutations.
思路：调用n-1次 leetcode 31 解题报告 中提到的算法即可。。。
代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月13日 星期四 14时49分34秒 4File Name :46.cpp 5 ************************************************ */ 6class Solution { 7 8 public: 9 10 void solve(vector<int> &nums) 11 { 12 int n = nums.size(); 13 if (n==0) return; 14 int k = -1; 15 for ( int i = n-2 ; i >= 0 ; i--) 16 { 17 if (nums[i]<nums[i+1]) 18 { 19 k = i ; 20 break; 21 } 22 } 23 if (k==-1) 24 { 25 reverse(nums.begin(),nums.end()); 26 return; 27 } 28 int l = -1; 29 for ( int i = n-1 ; i > k ; i-- ) 30 { 31 if (nums[k]<nums[i]) 32 { 33 l = i ; 34 break; 35 } 36 } 37 swap(nums[l],nums[k]); 38 reverse(nums.begin()+k+1,nums.end()); 39 } 40 vector<vector<int>> permute(vector<int>& nums) { 41 vector<vector<int> >res; 42 int n = nums.size(); 43 int total = 1; 44 for ( int i = 2 ; i <= n ; i++) total*=i; 45 for ( int i = 1 ; i <= total; i++) 46 { 47 res.push_back(nums); 48 solve(nums); 49 } 50 51 return res; 52 } 53 54};
Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
The replacement must be in-place, do not allocate extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column. 1,2,3 → 1,3,2 3,2,1 → 1,2,3 1,1,5 → 1,5,1
思路： 参考了wiki_Permutation