78. Subset
Topic link
Source: LeetCode
Link: https://leetcode-cn.com/problems/subsets
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Title description
Give you an integer array nums, the elements in the array are different from each other. Return all possible subsets (power sets) of this array.
The solution set cannot contain duplicate subsets. You can return the solution set in any order.
Example 1:
Input: nums = [1,2,3]
Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1, 2,3]]
Example 2:
Input: nums = [0]
Output: [[],[0]]
Topic analysis
1. Expansion method/violence method-one by one number expansion
Empty: []
Only 1: (add 1 to each subset) [1]
Add 2: (Add 2 to each subset) [2] [ 1,2]
plus 3: (add 3 to each subset) [3] [1, 3] [2, 3] [1, 2, 3]
No need to consider duplicate problems
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> res;
res.push_back(vector<int>());
for(int num : nums){
vector<vector<int>> subset;//存放每个数字产生新加的子集扩展 例如num为3 此时res【【】【1】【2】【1,2】】
for(auto vec: res){
vector<int> temp = vec;//temp为【】或【1】或【2】或【1,2】
temp.push_back(num);
subset.push_back(temp);
}
//此时subset为每个数产生的扩展 subset【【3】【1,3】【2,3】【1,2,3】】
for(auto s:subset){
res.push_back(s);
}
}
return res;
}
};
2. Backtracking recursion
length
0 【】
1 【1】【2】【3】
2 【1,2】【1,3】【2,3】
3 【1,2,3】
recursive downward each time The elements are the elements after the element
class Solution {
public:
vector<vector<int>> res;
vector<vector<int>> subsets(vector<int>& nums) {
res.push_back(vector<int>());
for(int length = 1; length <= nums.size(); length++){
//length是长度
backtracking(nums, length, 0, vector<int>());
}
return res;
}
void backtracking(vector<int>& nums, int length, int index, vector<int> temp){
if(temp.size() == length){
//剪枝
res.push_back(temp);
return ;
}
for(int i = index; i < nums.size(); i++){
temp.push_back(nums[i]);
backtracking(nums, length, i+1,temp);
temp.pop_back();
}
}
};
3.DFS depth-first algorithm-recursive (better to draw a picture) one way to the end
class Solution {
public:
vector<vector<int>> res;//2 3
vector<vector<int>> subsets(vector<int>& nums) {
dfs(nums, 0, vector<int>());//找到子集temp加入res index开始的索引位置
return res;
}
void dfs(vector<int>& nums, int index, vector<int> temp){
res.push_back(temp);
if(nums.size() == index){
return ;
}
for(int i = index; i < nums.size(); i++){
temp.push_back(nums[i]);
dfs(nums, i + 1, temp);
temp.pop_back();
}
}
};