zoukankan      html  css  js  c++  java
  • [leetcode-78-Subsets]

    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:
    [
    [3],
    [1],
    [2],
    [1,2,3],
    [1,3],
    [2,3],
    [1,2],
    []
    ]

    思路:

    首先排序,然后回溯法。

    void backtracking(vector<vector<int>>& subs,vector<int>& sub,vector<int>& nums,int start)
    {
      subs.push_back(sub);
      for(int i = start;i<nums.size();i++)
      {
        sub.push_back(nums[i]);
        backtracking(subs,sub,nums,i+1);
        sub.pop_back();
      }
    }
    vector<vector<int>> subsets(vector<int>& nums)
    {
      sort(nums.begin(),nums.end());
      vector<vector<int>> subs;
       vector<int>sub;
       backtracking(subs,sub,nums,0);   
      return subs;
    }

     另外,leetcode上大神有更详细多样的方法解析,如下:

    Recursive (Backtracking)

    This is a typical problem that can be tackled by backtracking. Since backtracking has a more-or-less similar template, so I do not give explanations for this method.

    class Solution {
    public:
        vector<vector<int>> subsets(vector<int>& nums) {
            sort(nums.begin(), nums.end());
            vector<vector<int>> subs;
            vector<int> sub;  
            genSubsets(nums, 0, sub, subs);
            return subs; 
        }
        void genSubsets(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]);
                genSubsets(nums, i + 1, sub, subs);
                sub.pop_back();
            }
        }
    };
    

    Iterative

    This problem can also be solved iteratively. Take [1, 2, 3] in the problem statement as an example. The process of generating all the subsets is like:

    1. Initially: [[]]
    2. Adding the first number to all the existed subsets: [[], [1]];
    3. Adding the second number to all the existed subsets: [[], [1], [2], [1, 2]];
    4. Adding the third number to all the existed subsets: [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]].

    Have you got the idea :-)

    The code is as follows.

    class Solution {
    public:
        vector<vector<int>> subsets(vector<int>& nums) {
            sort(nums.begin(), nums.end());
            vector<vector<int>> subs(1, vector<int>());
            for (int i = 0; i < nums.size(); i++) {
                int n = subs.size();
                for (int j = 0; j < n; j++) {
                    subs.push_back(subs[j]); 
                    subs.back().push_back(nums[i]);
                }
            }
            return subs;
        }
    }; 
    

    Bit Manipulation

    This is the most clever solution that I have seen. The idea is that to give all the possible subsets, we just need to exhaust all the possible combinations of the numbers. And each number has only two possibilities: either in or not in a subset. And this can be represented using a bit.

    There is also another a way to visualize this idea. That is, if we use the above example, 1 appears once in every two consecutive subsets, 2 appears twice in every four consecutive subsets, and 3 appears four times in every eight subsets, shown in the following (initially the 8 subsets are all empty):

    [], [], [], [], [], [], [], []

    [], [1], [], [1], [], [1], [], [1]

    [], [1], [2], [1, 2], [], [1], [2], [1, 2]

    [], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]

    The code is as follows.

    class Solution {
    public:
        vector<vector<int>> subsets(vector<int>& nums) {
            sort(nums.begin(), nums.end());
            int num_subset = pow(2, nums.size()); 
            vector<vector<int> > res(num_subset, vector<int>());
            for (int i = 0; i < nums.size(); i++)
                for (int j = 0; j < num_subset; j++)
                    if ((j >> i) & 1)
                        res[j].push_back(nums[i]);
            return res;  
        }
    };
    

    Well, just a final remark. For Python programmers, this may be an easy task in practice since the itertools package has a function combinations for it :-)

    参考:

    https://discuss.leetcode.com/topic/19110/c-recursive-iterative-bit-manipulation-solutions-with-explanations

  • 相关阅读:
    qt中使用C++thread
    qt--mask蒙版
    qt--调色板QPalette
    qt--Q_PROPERTY自定义属性
    路由扩张器
    qt-QBitmap单色图
    qt-QPixmap
    nodejs中的文件系统
    深入理解nodejs中的异步编程
    Vue 事件监听
  • 原文地址:https://www.cnblogs.com/hellowooorld/p/6938565.html
Copyright © 2011-2022 走看看