Subsets,Subsets II

最新推荐文章于 2022-05-07 14:28:24 发布

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原文链接：http://www.cnblogs.com/zengzy/p/5024597.html

本文详细介绍了两种子集生成算法：一种适用于不包含重复元素的整数集合；另一种则能够处理包含重复元素的情况。每种算法都通过递归深度优先搜索实现，并确保输出的子集按非递减顺序排列且不包含重复子集。通过具体实例展示了算法的运行结果。

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一.Subsets

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],
  []
]

class Solution {
public:
    void dfs(vector<int>& nums ,int numsSize,int startPos,vector<vector<int>>& res,vector<int>& oneOfRes)
    {
        for(int i=startPos;i<numsSize;i++){
            oneOfRes.push_back(nums[i]);
            res.push_back(oneOfRes);
            dfs(nums,numsSize,i+1,res,oneOfRes);
            oneOfRes.pop_back();
        }
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> res;
        vector<int> oneOfRes;
        res.push_back(vector<int>());
        int numsSize = nums.size();
        dfs(nums,numsSize,0,res,oneOfRes);
        return res;
    }
};

二.SubsetsII

Given a collection of integers that might contain duplicates, 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 = [1,2,2], a solution is:

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

class Solution {
public:
    void dfs(vector<int>& nums,int numsSize,int startPos,vector<vector<int>>& res,vector<int>& oneOfRes)
    {
        for(int i=startPos;i<numsSize;i++){
            if(i>startPos && nums[i]==nums[i-1]){
                continue;
            }
            oneOfRes.push_back(nums[i]);
            res.push_back(oneOfRes);
            dfs(nums,numsSize,i+1,res,oneOfRes);
            oneOfRes.pop_back();
        }
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> res;
        vector<int> oneOfRes;
        int numsSize = nums.size();
        res.push_back(oneOfRes);
        dfs(nums,numsSize,0,res,oneOfRes);
        return res;
    }
};