zoukankan      html  css  js  c++  java
  • [LeetCode] Subsets


    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.

     1 class Solution {
     2 public:
     3     vector<vector<int>> subsets(vector<int>& nums) {
     4         sort(nums.begin(), nums.end());
     5         vector<vector<int>> subs;
     6         vector<int> sub;
     7         genSubsets(nums, 0, sub, subs);
     8         return subs; 
     9     }
    10     void genSubsets(vector<int>& nums, int start, vector<int>& sub, vector<vector<int>>& subs) {
    11         subs.push_back(sub);
    12         for (int i = start; i < nums.size(); i++) {
    13             sub.push_back(nums[i]);
    14             genSubsets(nums, i + 1, sub, subs);
    15             sub.pop_back();
    16         }
    17     }
    18 };

    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.

     1 class Solution {
     2 public:
     3     vector<vector<int>> subsets(vector<int>& nums) {
     4         sort(nums.begin(), nums.end());
     5         vector<vector<int>> subs(1, vector<int>());
     6         for (int i = 0; i < nums.size(); i++) {
     7             int n = subs.size();
     8             for (int j = 0; j < n; j++) {
     9                 subs.push_back(subs[j]); 
    10                 subs.back().push_back(nums[i]);
    11             }
    12         }
    13         return subs;
    14     }
    15 };

    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, 1appears 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 8subsets 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.

     1 class Solution {
     2 public:
     3     vector<vector<int>> subsets(vector<int>& nums) {
     4         sort(nums.begin(), nums.end());
     5         int num_subset = pow(2, nums.size()); 
     6         vector<vector<int> > res(num_subset, vector<int>());
     7         for (int i = 0; i < nums.size(); i++)
     8             for (int j = 0; j < num_subset; j++)
     9                 if ((j >> i) & 1)
    10                     res[j].push_back(nums[i]);
    11         return res; 
    12     }
    13 };

    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 :-) 

  • 相关阅读:
    jquery属性
    jquery选择器
    Django的模型
    win7安装RabbitMQ
    阿里云RDS备份的tar格式包恢复到本地自建数据库
    正确使用 Volatile 变量
    深入分析Volatile的实现原理
    volatile和synchronized的区别
    全面理解Java内存模型
    深入理解Feign之源码解析
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4548112.html
Copyright © 2011-2022 走看看