LeetCode 78. Subsets

最新推荐文章于 2022-10-22 22:28:31 发布

原创最新推荐文章于 2022-10-22 22:28:31 发布 · 253 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解同时被 2 个专栏收录

288 篇文章

订阅专栏

Multiple Passes

85 篇文章

订阅专栏

本文详细介绍了使用递归生成子集的方法，包括使用深度优先搜索（DFS）、位操作等技巧，通过实例展示了如何从一组不同的整数中生成所有可能的子集，并对关键步骤进行了注释和说明。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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],
  []
]

There are several ways to solve it.... dfs, bfs, bit-manipulation... and so on.

DFS as follows.

// Given a set of distinct integers, nums, return all possible subsets.

#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

void subsets(vector<int>& nums, int start, vector< vector<int> >& res, vector<int>& path) {
    res.push_back(path);
    for(int i = start; i < nums.size(); ++i) {
        path.push_back(nums[i]);
        subsets(nums, i + 1, res, path);
        path.pop_back();
    }
}

vector< vector<int> > subsets(vector<int>& nums) {
    vector< vector<int> > res;
    vector<int> path;
    sort(nums.begin(), nums.end());
    subsets(nums, 0, res, path);
    return res;
}


vector< vector<int> > subsetsII(vector<int>& nums) {
    vector< vector<int> > res;
    vector<int> path;
    sort(nums.begin(), nums.end());
    function<void(int)> backtrack = [&](int start) {
        res.push_back(path);
        for(int i = start; i < nums.size(); ++i) {
            path.push_back(nums[i]);
            backtrack(i+1);
            path.pop_back();
        }                                             //c++11 support inner function definition....exciting!!
    };
    backtrack(0);
    return res;
}<pre name="code" class="cpp">int main(void) {
    vector<int> nums;
    nums.push_back(1);
    nums.push_back(2);
    nums.push_back(3);
    vector< vector<int> > res = subsets(nums);
    for(int i = 0; i < res.size(); ++i) {
        for(int j = 0; j < res[i].size(); ++j) {
            cout << res[i][j] << endl;
        }
        cout << endl;
    }
}

vector< vector<int> > subsetIterative(vector<int>& array) {
  int n = array.size();
  int m = pow(2, n);
  vector< vector<int> > res(m, vector<int>());
  for(int i = 0; i < m; ++i) {
    for(int j = 0; j < n; ++j) {
      if(i & (1 << j)) res[i].push_back(array[j]);
    }
  }
  return res;
}