算法总结(7)--leetcode上的递归,BFS,DFS思考

recursion
n. 递推; 递归，递归式;

下面的题目，也不是很难，不过需要注意递归调用的条件，如何递归
还有就是题意的理解

另外设计到重复结果的处理，如果序列有序的，可以在遍历的时候就处理，减少DFS数，还有就是直接用set去存储结果，保证不重复

---

38. Count and Say

题目地址

https://leetcode.com/problems/count-and-say/

题目意思和参考

http://blog.csdn.net/fly_yr/article/details/48000713

递归ac代码

class Solution {public:    string countAndSay(int n) {        string rs;        if (n <= 0)            return rs;        if (n == 1)            return "1";        string ans = countAndSay(n - 1); //递归        int len = ans.size();        int cnt = 1;        for (int i = 0; i < len; i++)        {            if (i == len - 1)            {                rs += to_string(cnt);                rs += ans[i];                break;            }            if(ans[i] == ans[i+1])            {                cnt++;            }            else{                rs += to_string(cnt);                rs += ans[i];                cnt = 1;            }        }        return rs;    }};

22. Generate Parentheses

题目地址

https://leetcode.com/problems/generate-parentheses/

题目描述

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

[  "((()))",  "(()())",  "(())()",  "()(())",  "()()()"]

求解

这题采用的是dfs 回溯的方法，主要是如何构建递归函数

ac代码如下

class Solution {public:    vector<string> ans;    /*        left代表左括号的数目 <= n        depth代表总共的数目  <= 2*n    */    void dfs(string s, int left,int depth,int n)    {        if (depth >= 2 * n)        {            //cout << s << endl;            ans.push_back(s);            return;        }        if (left == n)        {            string str = s;            for (int i = 0; i < 2*n - depth;i++)            {                s += ')';            }            dfs(s, n, 2*n, n);            s = str;        }        else if (depth == 2 * left){ //左右括号都是比配的，只能加左括号了            string str = s;            s += "(";            dfs(s, left + 1, depth + 1, n);            s = str;        }        else{            string str = s;            s += "(";            dfs(s, left + 1, depth + 1, n);            s = str;            str = s;            s += ")";            dfs(s, left, depth + 1, n);            s = str;        }    }    vector<string> generateParenthesis(int n) {        string s = "(";        dfs(s, 1,1, n);        return ans;    }};

或

class Solution {public:    vector<string> ans;    void dfs(string &s , int leftCnt, int totalCnt, int n)    {        if(leftCnt > n || totalCnt > 2*n)            return;        int rightCnt = totalCnt - leftCnt;        if(rightCnt > leftCnt)            return;        if(leftCnt == n && totalCnt == 2*n)        {            ans.push_back(s);            //cout << s << endl;            return;        }        if(leftCnt < n)        {            string tmp = s;            s += ")";            dfs(s, leftCnt, totalCnt + 1, n);            s = tmp + "(";            dfs(s, leftCnt + 1, totalCnt + 1, n);        }else{            s += ")";            dfs(s, leftCnt, totalCnt + 1, n);        }    }    vector<string> generateParenthesis(int n) {        string s = "(";        dfs(s, 1, 1, n);        return ans;    }};

279. Perfect Squares

题目地址

https://leetcode.com/problems/perfect-squares/

题目描述

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, …) which sum to n.

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

ac (bfs 记忆)

class Solution {public:    int bfs(vector<int> &nums, int len , int n)    {        vector<bool> vis(n+1,false);        queue<int> que;        queue<int> queCeng;        que.push(0);        queCeng.push(0);        vis[0] = true;        while(!que.empty())        {            int curSum = que.front();            que.pop();            int curCeng = queCeng.front();            queCeng.pop();            if(curSum == n){                return curCeng;            }            for(int i=0;i<len;i++)            {                if(curSum + nums[i] <= n && vis[curSum + nums[i]] == false)                {                    vis[curSum + nums[i]] = true;                    que.push(curSum + nums[i]);                    queCeng.push(curCeng + 1);                }            }        }        return -1;    }    int numSquares(int n) {        vector<int> nums;        int len = 0;        int i = 1;        while(i * i <= n)        {            nums.push_back(i*i);            len ++;            i++;        }        if(nums[len - 1] == n)        {            return 1;        }        return bfs(nums, len, n);    }};

491. Increasing Subsequences

题目地址

https://leetcode.com/problems/increasing-subsequences/

ac

排了序的话
是可以按照
/*if(i > index && num[i] == num[i-1])
continue;*/
但如果本身就是无序的话，应该是不可以的，仍然会有重复的情况出现

DFS ac代码

class Solution {public:    vector<vector<int>> ans;    vector<int> rs;    set<vector<int>> se;    vector<int> num;    int len;    void dfs(int curNum, int index,int curLen)    {        if(curLen >= 2)        {            //ans.push_back(rs);            se.insert(rs);        }        for(int i= index ;i < len;i++)        {            /*if(i > index && num[i] == num[i-1])                continue;*/            if(num[i] >= curNum)            //if(rs.size() == 0 || num[i] >= rs[rs.size() - 1])            {                rs.push_back(num[i]);                dfs(num[i], i+1, curLen + 1);                rs.pop_back();            }        }    }    vector<vector<int>> findSubsequences(vector<int>& nums) {        len = nums.size();        num = nums;        if(len < 2)            return ans;        int maxx = 0x7fffffff;        int minn = -1 - maxx;        dfs(minn, 0, 0);        for(auto it=se.begin();it!=se.end();it++)             ans.push_back(*it);          return ans;    }};

---

#