Subsets I:
Given a set of distinct integers, nums, return all possible subsets.
Note:
- Elements in a subset must be in non-descending order.
- The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,3], a solution is:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
New Version: we can find that if we have the subsets of N-1 numbers, then the result is composed of adding the Nth number to each of the subsets of N-1 numbers plus the original subsets of N-1 numbers, i.e. S(N) = S(N-1) + (adding N to S(N-1)).
Runtime: 4ms
1 class Solution { 2 public: 3 vector<vector<int>> subsets(vector<int>& nums) { 4 vector<vector<int> > result; 5 if(nums.empty()) return result; 6 7 sort(nums.begin(), nums.end()); 8 vector<int> temp; 9 10 //manipulate the first number in nums 11 result.push_back(temp); 12 temp.push_back(nums[0]); 13 result.push_back(temp); 14 int level = 1; 15 16 //manipulate the remaining numbers in nums 17 while(level < nums.size()){ 18 int n = result.size(); 19 for(int i = 0; i < n; i++){ 20 temp = result[i]; 21 temp.push_back(nums[level]); 22 result.push_back(temp); 23 } 24 level++; 25 } 26 27 return result; 28 } 29 };
Analyse: Each number has two states: selected or not selected. Recursively invoke the choose function until the step equals to the size. Then construct a vector to store all selected numbers, and push it to the result vector.
Runtime: 8ms.
1 class Solution { 2 public: 3 vector<vector<int>> subsets(vector<int>& nums) { 4 vector<vector<int> > result; 5 sort(nums.begin(), nums.end()); 6 vector<bool> selected(nums.size(), false); 7 subset(nums, selected, 0, result); 8 9 return result; 10 } 11 12 void subset(vector<int> &nums, vector<bool> &selected, int step, vector<vector<int> > &result){ 13 if(step == nums.size()){ 14 vector<int> temp; 15 for(int i = 0; i < step; i++){ 16 if(selected[i]) temp.push_back(nums[i]); 17 } 18 result.push_back(temp); 19 return; 20 } 21 22 //not choose the number 23 selected[step] = false; 24 subset(nums, selected, step + 1, result); 25 //choose the number 26 selected[step] = true; 27 subset(nums, selected, step + 1, result); 28 } 29 };
Subsets II:
Given a collection of integers that might contain duplicates, nums, return all possible subsets.
Note:
- Elements in a subset must be in non-descending order.
- The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,2], a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
Analyse: It is the same as subsets I except that we need to set a variable keep the place where we start to push the last number.
Runtime: 12ms
1 class Solution { 2 public: 3 vector<vector<int>> subsetsWithDup(vector<int>& nums) { 4 vector<vector<int> > result; 5 if(nums.empty()) return result; 6 7 sort(nums.begin(), nums.end()); 8 vector<int> temp; 9 10 //manipulate the first number in the nums 11 result.push_back(temp); 12 temp.push_back(nums[0]); 13 result.push_back(temp); 14 int level = 1; 15 int index = 1; //the place where push the first number 16 17 //manipulate remaining numbers in the nums 18 while(level < nums.size()){ 19 int i; 20 if(nums[level] == nums[level - 1]) 21 i = index; 22 else i = 0; 23 index = result.size(); 24 int n = result.size(); 25 for(; i < n; i++){ 26 temp = result[i]; 27 temp.push_back(nums[level]); 28 result.push_back(temp); 29 } 30 level++; 31 } 32 return result; 33 } 34 };