78 Subsets

最新推荐文章于 2020-12-08 16:01:21 发布

weixin_39145266

最新推荐文章于 2020-12-08 16:01:21 发布

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分类专栏： # Array # Solved LeetCode # Medium # Backtracking # BitManipulation

本文链接：https://blog.youkuaiyun.com/weixin_39145266/article/details/88850167

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本文介绍了一种生成给定整数数组所有可能子集(幂集)的算法，并提供了两种不同的实现方式：一种采用递归方法逐步构建子集，另一种通过迭代在每一步选择或不选择元素来生成子集。这两种方法都能确保结果中不含重复子集。

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1 题目

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3]
Output:
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

2 尝试解

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<int> temp;
        vector<vector<int>> result;
        if(nums.size() == 0) {
            result.push_back(temp);
            return result;
        }
        int index = 0;
        getSubset(nums,index,result,temp);
        return result;
    }
    void getSubset(vector<int>&nums, int index, vector<vector<int>>& result,vector<int> &temp){
        if(index == nums.size()-1) {
            temp.push_back(nums[index]);
            result.push_back(temp);
            temp.pop_back();
            result.push_back(temp);
            return;
        }
        temp.push_back(nums[index]);
        getSubset(nums,index+1,result,temp);
        temp.pop_back();
        getSubset(nums,index+1,result,temp);
    }
};

3 标准解

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> subs;
        vector<int> sub;
        subsets(nums, 0, sub, subs);
        return subs;
    }
private:
    void subsets(vector<int>& nums, int i, vector<int>& sub, vector<vector<int>>& subs) {
        subs.push_back(sub);
        for (int j = i; j < nums.size(); j++) {
            sub.push_back(nums[j]);
            subsets(nums, j + 1, sub, subs);
            sub.pop_back();
        }
    }
};