递归那些事儿

这个里面用来写一些我遇到的递归题目

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], [] ]

这是一个比较典型的递归问题

 vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> ans;
        vector<int> res;
        sort(nums.begin(),nums.end());
        dfs(0,res,nums,ans);
        return ans;
    }
    void dfs(int cur,vector<int>&res,vector<int>& nums,vector<vector<int>>& ans){
        if(cur==nums.size()){//整个nums都被遍历
            ans.push_back(res);
            return;
        }
        int oldsize=res.size();
        res.push_back(nums[cur]);
        dfs(cur+1,res,nums,ans);
        res.resize(oldsize);
        dfs(cur+1,res,nums,ans);
    }

除此之外，这个题目也可以使用DP来完成

 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;
    }

再者还有另外一种解法就是使用bit串的形式
这真的是一种很漂亮的解法

 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;  
    }

---

Word Search

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent
cell, where “adjacent” cells are those horizontally or vertically
neighboring. The same letter cell may not be used more than once.

For example, Given board =

[ [“ABCE”], [“SFCS”], [“ADEE”] ] word = “ABCCED”, -> returns
true, word = “SEE”, -> returns true, word = “ABCB”, -> returns false.

这个题目需要注意的是，同一个位置不可以遍历两遍。所以需要一个res数组来记录当前位置是否已经走过。

bool bt(vector<vector<char>>& board,vector<vector<int>>&rec,string word,int i,int j){
        if(word.size()==0)
            return true;
        if(i>board.size()-1||i<0||j>board[0].size()-1||j<0)
            return false;
        if(word[0]!=board[i][j]||rec[i][j])
            return false;
        rec[i][j]=1;
        string ns=word.substr(1,word.size()-1);
        bool res= bt(board,rec,ns,i-1,j)||bt(board,rec,ns,i+1,j)||bt(board,rec,ns,i,j-1)||bt(board,rec,ns,i,j+1);
        rec[i][j]=0;
        return res;
    }

bool exist(vector<vector<char>>& board, string word) {
        vector<vector<int>> rec(board.size(),vector<int>(board[0].size(),0));

        for(int i=0;i<board.size();i++)
            for(int j=0;j<board.size();j++)
                if(bt(board,rec,word,i,j))
                    return true;
        return false;
    }

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