递增子序列

class Solution {
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
       
        set<vector<int>> st;//用 set 保存结果去重复
        vector<int> cur;//当前子序列
        dfs(st,cur,nums,0);
        vector<vector<int>> res(st.begin(),st.end());  
        return res;
        
    }

    void dfs(set<vector<int>> &st, vector<int> &cur,vector<int>& nums,int curIndex )
    {
        if( curIndex >= nums.size() ) //超边界，没元素可选
        {
            //判断当前子序列
            if( cur.size() > 1)
                st.insert(cur);
            return;
        }
        
        //遍历当前所有可能的选择：选择curIndex 、不选curIndex
        //选择curIndex
        if( cur.size() == 0 || nums[curIndex] >= cur.back())
        {
            cur.push_back(nums[curIndex]);//可以加入序列
            dfs( st,cur,nums,curIndex+1);
            cur.pop_back();//撤销选择
        }
        //不选curIndex
        dfs( st,cur,nums,curIndex+1);
    }
       
};

标签：cur,nums,递增,dfs,st,vector,序列,curIndex
来源： https://blog.csdn.net/jwy2014/article/details/115271535