[算法专题] 深度优先搜索&回溯剪枝

1. Palindrome Partitioning

https://leetcode.com/problems/palindrome-partitioning/

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]
/**
 * author : Jianxin Zhou
 * email:zhoujx0219@163.com
 * 
 * 该题dfs函数原型如下:
 * void partitionHelper(const string &s, vector<vector<string>> &result, vector<string> &path, int pos)
 * 
 * 以aaba举例。
 * 1. 首先a为回文,然后对aba进行dfs
 * 2. 之后回溯到a时,以aa为回文,然后对ba做dfs
 * 3. 回溯到aa,试图以aab为回文,失败;试图以aaba为回文失败;结束。
 * 
 * 注意:如果能顺利的找到一组回文,那么pos最终会等于s.size(),此时可以push到result。
 *       如果找不到,例如之前的aaba不是回文,那么就会直接退出循环,没有机会执行下一步递归,也就没有pos等于s.size了。
 * 
 * 实际上,此类题与真正的dfs的差别在于,dfs在回溯时,不会进行剪枝操作。而此类题,由于需要求出所有方案,所以需要剪枝。
 *
 */


class Solution {
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> result;
        vector<string> path;
        partitionHelper(s, result, path, 0);
        return result;
    }
    
private:
    void partitionHelper(const string &s, vector<vector<string>> &result, vector<string> &path, int pos) {
        // base case
        if (pos == s.size()) {
            result.push_back(path);
            return;
        }
        
        for (int i = pos; i < s.size(); i++) {
            if (isPalindrome(s, pos, i)) {
                path.push_back(s.substr(pos, i - pos + 1));
                partitionHelper(s, result, path, i + 1);
                path.pop_back();
            }
        }
    }
    
    bool isPalindrome(const string &s, int start, int end) {
        while (start < end) {
            if (s[start] == s[end]) {
                start++;
                end--;
            } else {
                break;
            }
        }
        
        return start >= end;
    }
};

2. Permutations

https://leetcode.com/problems/permutations/

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

具体可参加我之前写的文章:[LintCode] Permutations

/**
 * 思路:dfs。
 * 
 * 以123举例,
 * 1. 首先以1作为head,然后对23做dfs
 * 2. 回溯到1, 以2作为head,对13做dfs
 * 3. 最后回溯到2,以3作为head,对12做dfs
 * 
 * 注意:例如以2为head,对其余元素做dfs时,那么2不能再取,因此在进行下一轮dfs时,需要标记2为以访问过
 * 
 */


class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> result;
        vector<int> path;
        
        bool visited[nums.size()];
        for(int i = 0; i < nums.size(); i++) {
            visited[i] = false;
        }
        
        sort(nums.begin(), nums.end());
        dfs(nums, result, path, visited);
        return result;
    }
    
private:
    void dfs(const vector<int> &nums, vector<vector<int>> &result, vector<int> &path, bool visited[]) {
        // base case
        if (path.size() == nums.size()) {
            result.push_back(path);
            return;
        }

        for (int i = 0; i < nums.size(); i++) {
            if (visited[i] == false) {
                path.push_back(nums[i]);
                visited[i] = true;
                dfs(nums, result, path, visited);
                path.pop_back();
                visited[i] = false;
            }
            
        }
    }
};

3. Permutations II

https://leetcode.com/problems/permutations-ii/

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

要点在于保证相同的数不在同一位置出现两次以上,可以参见我写的这篇文章:[LintCode] Permutations II

class Solution {
public:
    /**
     * @param nums: A list of integers.
     * @return: A list of unique permutations.
     */
    vector<vector<int> > permuteUnique(vector<int> &nums) {
        // write your code here
        vector<vector<int>> paths;
        if (nums.empty()) {
            return paths;
        }
         
        sort(nums.begin(), nums.end());
        bool *visited = new bool[nums.size()]();
        vector<int> path;
        permuteUniqueHelper(nums, visited, path, paths);
        return paths;
    }
     
private:
    void permuteUniqueHelper(const vector<int> &nums,
                             bool visited[],
                             vector<int> &path,
                             vector<vector<int>> &paths) {
        if (path.size() == nums.size()) {
            paths.push_back(path);
            return;
        } 
         
        for (int ix = 0; ix < nums.size(); ix++) {
            if (visited[ix] == true || ix > 0 && nums[ix - 1] == nums[ix] && visited[ix - 1] == false) {
                continue;    
            }
             
            visited[ix] = true;
            path.push_back(nums[ix]);
            permuteUniqueHelper(nums, visited, path, paths);
            visited[ix] = false;
            path.pop_back();
        }
    }
};

4 Subsets

https://leetcode.com/problems/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],
  []
]
/**
 * 思路:找方案,一般都是使用搜索。
 * 
 * 以123为例,在递归还没有开始前,先把空集push到result中,之后:
 * 1. 以1位head,对23做dfs,所以pos需要加1,用于分支限界(1 12 13 123)
 * 2. 回溯到1,以2为head,对3做dfs (2 23)
 * 3. 回溯到3,以3为head,之后循环结束。 (3)
 * 
 * 
 */



class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        // ensure that elements in a subset must be in non-descending order.
        sort(nums.begin(), nums.end());
        
        vector<vector<int>> res;
        vector<int> path;
        dfs(nums, res, path, 0);
        return res;
    }
    
private:
    void dfs(const vector<int> &nums, vector<vector<int>> &res, vector<int> &path, int pos) {
        res.push_back(path);
        
        for (int i = pos; i < nums.size(); i++) {
            path.push_back(nums[i]);
            dfs(nums, res, path, i + 1);
            path.pop_back();
        }
    }
};

5. Subsets II

https://leetcode.com/problems/subsets-ii/

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],
  []
]

同一位置上,前面取过的数,后面就不要再重复取了,当然当i = pos时,这个数必然是第一次取。

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

6 Restore IP Addresses

https://leetcode.com/problems/restore-ip-addresses/

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given "25525511135",

return ["255.255.11.135", "255.255.111.35"]. (Order does not matter)

/**
 * 该题思路与求回文划分相似
 */

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> res;
        
        size_t len = s.size();
        if (len < 4 || len > 12) {
            return res;
        }
        

        vector<string> path;
        dfs(s, res, path, 0);
        return res;
    }
    
private:
    void dfs(const string &s, vector<string> &res, vector<string> &path, int pos) {
        // base case
        if (path.size() == 4) {
            if (pos != s.size()) {
                return;
            }
            
            string returnElem;
            for (const auto &elem : path) {
                returnElem += elem;
                returnElem += ".";
            }
            returnElem.erase(returnElem.end() - 1);
            
            res.push_back(returnElem);
            return;
        }
        
        for (int i = pos; i < s.size() && i < pos + 3; i++) {
            string tmp = s.substr(pos, i - pos + 1);
            if (isValid(tmp)) {
                path.push_back(tmp);
                dfs(s, res, path, i + 1);
                path.pop_back();
            }
        }
    }
    
    bool isValid(const string &s) {
        // 排除 055 之类的数字
        if (s[0] == '0' && s.size() > 1) {
            return false;
        }
        
        int digit = atoi(s.c_str());
        return 0 <= digit && digit <= 255;
    }
};

7 N-Queens

http://www.lintcode.com/en/problem/n-queens/#

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

image

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]
/**
 * 思路:一行一行的取数,例如第一行的皇后放在第1个位置,第二行的皇后放在第3个位置,
 * 以此类推,直到最后一行的皇后放在正确的位置,如此视为一个方案,push到result中
 * 
 * 显然,本题使用dfs,每一行可取的位置从0-N-1,
 * 需要注意的是,每一行在取位置的时候,需要判断有效性(是否可以相互攻击)。
 */



class Solution {
public:
    /**
     * Get all distinct N-Queen solutions
     * @param n: The number of queens
     * @return: All distinct solutions
     * For example, A string '...Q' shows a queen on forth position
     */
    vector<vector<string> > solveNQueens(int n) {
        vector<vector<string>> res;
        vector<int> visitedCol;
        
        if (n <= 0) {
            return res;
        }
        
        dfs(n, res, visitedCol);
        return res; 
    }
    
private:
    void dfs(const int n, vector<vector<string>> &res, vector<int> &visitedCol) {
        // base case
        if (visitedCol.size() == n) {
            res.push_back(draw(visitedCol));
            return;
        }
        
        for (int i = 0; i < n; i++) {
            if (!isValid(visitedCol, i)) {
                continue;
            }
            
            visitedCol.push_back(i);
            dfs(n, res, visitedCol);
            visitedCol.pop_back();
        }
        
        
    }
    
    bool isValid(const vector<int> &visitedCol, const int currentCol) {
        size_t currentRow = visitedCol.size();
        
        for (int rowIndex = 0; rowIndex < visitedCol.size(); rowIndex++) {
            if (currentCol == visitedCol[rowIndex]) {
                return false;
            }
            
            if (currentRow + currentCol == rowIndex + visitedCol[rowIndex]) {
                return false;
            }
            
            if (currentRow - currentCol == rowIndex - visitedCol[rowIndex]) {
                return false;
            }
        }
        
        return true;
    }
    
    vector<string> draw(const vector<int> &visitedCol) {
        
        vector<string> ret;
        string row;
        for (const auto &elem : visitedCol) {
            row.clear();
            
            for (int i = 0; i < visitedCol.size(); i++) {
                if (i == elem) {
                    row += "Q";
                } else {
                    row += ".";
                }
            }
            
            ret.push_back(row);
        }
        
        return ret;
    }
};

8 Sudoku Solver

https://leetcode.com/problems/sudoku-solver/

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

...and its solution numbers marked in red.

class Solution {
public:
    void solveSudoku(vector<vector<char>>& board) {
        dfs (board, 0, 0);
    }
    
private:
    /**
     * 该题需要对sudoku中每一个以‘.’标记的方格进行dfs,
     * 1. 如果对当前方格的以1-9这9个数字进行遍历,都不合法,那么不会再往下一个方格进行dfs,直接回溯到上一个方格取下一个数。
     * 2. 如果当前方格所取的数合法,那么继续对下一个方格进行dfs,依次下去如果一直合法,那么直到走到sudoku中的最后一个需要放数字的方格,
     *    尝试完它的所有选择,再往上回溯。
     * 然后,在这边我们只需要一个可行解即可,因此只要当前方格合法,往下的dfs返回true,那么即为一个解,直接返回。
     * 
     * 
     * 
     */
    bool dfs(vector<vector<char>> &board, int x, int y) {

        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                //dfs
                if (board[i][j] == '.') {
                    // k从0-9走完才算走完,但是此处我们只要有一个解,就可以返回了,因此在以下循环中设置了return语句
                    for (int k = 0; k < 9; k++) {
                        bool flag;
                        
                        if (!isValid(board, i ,j, k)) {
                            continue;
                        }
                        
                        board[i][j] = '1' + k;
                        
                        if (j != 8) {
                            flag = dfs(board, i, j + 1);
                        } else {
                            flag = dfs(board, i + 1, 0);
                        }
                        
                        // 当前合法&&下一轮dfs合法,说明找到解
                        if (flag) {
                            return true;
                        }
                        
                        board[i][j] = '.';
                    }
                    
                    // 遍历完9个数,仍然找不到合适的解,则返回false
                    return false;
                }
            }
        }
        
        // 当所有各自都走完,自然返回true(注意只有当前合法,才会继续往下走,继续往下走的最终结果是越了sudoku的界限)
        return true;
        
    }
    
    bool isValid(const vector<vector<char>> &board, int x, int y, int k) {
        int i, j;
        for (i = 0; i < 9; i++) // 检查 y 列
            if (i != x && board[i][y] == '1' + k)
                return false;
        
        for (j = 0; j < 9; j++) // 检查 x 行
            if (j != y && board[x][j] == '1' + k)
                 return false;

        for (i = 3 * (x / 3); i < 3 * (x / 3 + 1); i++)
            for (j = 3 * (y / 3); j < 3 * (y / 3 + 1); j++)
                if ((i != x || j != y) && board[i][j] == '1' + k)
                    return false;

        return true;
    }
};
 

小结1

做搜索的题目,最关键的是要知道对什么对象进行dfs,例如,在sudoku中是对每一个以“.”标记的方格进行dfs,在回文划分中,是对每一个划分的位置进行dfs,在8妃问题中,是对每一行妃子可以在的位置进行dfs。

其次,dfs时,我们需要判断所取的每一个解是否是有效的,最好写一个函数来专门做这件事情。只要当当前对象dfs的数值有效时,才会继续往对下一个对象进行dfs,否则就直接向上回溯了(这点可以参见sudoku中的解释)。

最后,对于每次dfs时,可以对范围进行分支限界。例如回文划分、subset等。

小结2

值得注意的是:到底要对多少对象进行dfs,有时候是很明显的,例如8妃和sudoku问题,8妃就是对8行依次dfs,sudoku就是对所有方格进行dfs。但有时,总共要对多少对象进行dfs并不明显。dfs的递归基要处理的就是dfs完多少个对象就一定要返回(不然就无限dfs下去了)。当然,在sudoku问题中,方格的循环走完返回,这是一个隐含的递归基。

总结:dfs函数中,递归基处理的是dfs多少个对象就要返回。而每次dfs的for循环,往往是每一次dfs的范围。当递归栈最顶层的那个dfs循环走完,搜素就完成了。

小结3

在图论中,往往是从某一个点开始往下dfs,dfs的范围是当前node的所有neighbor,与我们通常的搜索问题不同的是,图论中的dfs在回溯时不会剪枝,总之,找到一条路径就结束了。

 

 

posted @ 2015-08-07 21:59  Acjx  阅读(3492)  评论(2编辑  收藏  举报