LeetCode 78. Subsets
原题目:
Given a set of distinct integers, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,3], a solution is:
[
[],
[],
[],
[,,],
[,],
[,],
[,],
[]
]
题目翻译:
给出一个不含重复元素的整型数组,找到所有该数组的子集,返回的结果不能包含重复的子集。
思路:
这是一道经典的子集树问题,可以直接用回溯法(Backtracking)解决(也可称之为递归法,Recursive)。查看答案之后发现,还有另外两种方案可以选择,三种方案分别是:回溯(Backtracking),迭代(Iterative),位操作(Bit Manipulation)。在此做一个整理。
方案一:回溯(Backtracking)
根据深度优先搜索(DFS)的思想实现。
AC代码如下,(C++,6ms)
class Solution {
public:
void backtracking(vector<int>& nums, int start, vector<int>& sub, vector<vector<int>>& subs) {
subs.push_back(sub);
for (int i = start; i < nums.size(); i++) {
sub.push_back(nums[i]);
backtracking(nums, i + , sub, subs);
sub.pop_back();
}
}
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> subs;
vector<int> sub;
backtracking(nums, , sub, subs);
return subs;
}
};
方案二:迭代(Iterative)
因为n个元素的集合一共有 2n 个子集(包括空集和集合本身),所以据此可以得到一个子集的生成策略,以[1, 2, 3]为例:
0)初始结果为只包含空集的集合:
subs = [ [] ];
1)将结果里的所有集合复制一次:
subs = [ [], [] ];
把第一个元素放入所有新生成的集合的尾部:
subs = [ [], [1] ];
2)重复步骤1,放入第二个元素,边复制边放入:
subs = [ [], [1], [] ];
subs = [ [], [1], [2] ];
subs = [ [], [1], [2], [1] ];
subs = [ [], [1], [2], [1, 2] ];
3)重复步骤1,放入第三个元素:
subs = [ [], [1], [2], [1, 2], [] ];
subs = [ [], [1], [2], [1, 2], [3] ];
subs = [ [], [1], [2], [1, 2], [3], [1] ];
subs = [ [], [1], [2], [1, 2], [3], [1, 3] ];
subs = [ [], [1], [2], [1, 2], [3], [1, 3], [2] ];
subs = [ [], [1], [2], [1, 2], [3], [1, 3], [2, 3] ];
subs = [ [], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2] ];
subs = [ [], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3] ];
AC代码如下,(C++,6ms)
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> subs(, vector<int>());
for (int i = ; i < nums.size(); i++) {
int n = subs.size();
for (int j = ; j < n; j++) {
subs.push_back(subs[j]);
subs.back().push_back(nums[i]);
}
}
return subs;
}
};
方案三:位操作(Bit Manipulation)
这个方案也是利用了n个元素的集合一共有 2n 个子集的性质,对集合中的每一个元素,将其插入所在的子集里面。还是以[1, 2, 3]为例:
0)初始结果为包含8个空集的集合( 23=8 ):
subs = [ [], [], [], [], [], [], [], [] ];
1)将第一个元素插入,隔 20=1 个插入连续的1个:
subs = [ [], [1], [], [1], [], [1], [], [1] ];
2)将第二个元素插入,隔 21=2 个插入连续的2个:
subs = [ [], [1], [2], [1, 2], [], [1], [2], [1, 2] ];
3)将第三个元素插入,隔 22=4 个插入连续的4个:
subs = [ [], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3] ];
4)由于其中多处的值是2的幂次方,因此可以使用位运算的移位操作和与操作提高效率。
AC代码如下,(C++,3ms)
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
int num_subset = << nums.size();
vector<vector<int> > subs(num_subset, vector<int>());
for (int i = ; i < nums.size(); i++)
for (int j = ; j < num_subset; j++)
if ((j >> i) & )
subs[j].push_back(nums[i]);
return subs;
}
};