LeetCode专题-深度优先搜索(二)

目录

N皇后问题一、二

51. N-Queens

Hard

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

八皇后

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.

Example:

Input: 4
Output: [
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.

题目大意：在nxn的棋盘上放置n个国际象棋的皇后，要求皇后之间不能够相互攻击到。

解题思路：利用DFS，从第一行开始，放置皇后在安全的位置。

class Solution {
public:
    // check if put a queen at [row, col] is valid
    bool isValid(const vector &cur, int row, int col){
        // check if queen exists in the same col
        for(int i = 0; i < row; i++){
            if(cur[i][col] == 'Q'){
                return false;
            }
        }

        // check if queen exists in up-left diag
        for(int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--){
            if(cur[i][j] == 'Q'){
                return false;
            }
        }

        // check if queen exists in up-right diag
        for(int i = row - 1, j = col + 1; i >= 0 && j < cur.size(); i--, j++){
            if(cur[i][j] == 'Q'){
                return false;
            }
        }

        return true;
    }

    void solveNQueensDFS(vector &cur, int row, vector> &ans){
        if(row == cur.size()){
            //if all queens have been positioned
            ans.push_back(cur);
            return;
        }

        // try to place queen at one col of current row
        for(int col = 0; col < cur.size(); col++){
            if(isValid(cur, row, col)){
                cur[row][col] = 'Q';
                solveNQueensDFS(cur, row + 1, ans);
                cur[row][col] = '.';
            }
        }
    }

    vector> solveNQueens(int n) {
        vector> ans;
        vector  cur(n, string(n, '.'));
        solveNQueensDFS(cur, 0, ans);
        return move(ans);
    }
};

测试一下

Success
Details
Runtime: 8 ms, faster than 92.21% of C++ online submissions for N-Queens.
Memory Usage: 10.1 MB, less than 66.69% of C++ online submissions for N-Queens.

52. N-Queens II

Hard

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

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

Example:

Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],

["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]

题目大意：和上题相比稍加修改，要求求出可能的解法个数。

解题思路：思路与上题相同。

class Solution {
public:
    int totalNQueens(int n) {
        int cnt = 0;
        vector  cur(n, string(n, '.'));
        solveNQueensDFS(cur, 0, cnt);
        return cnt;        
    }
    
    bool isValid(const vector &cur, int row, int col){
        for(int i = 0; i < row; i++){
            if(cur[i][col] == 'Q'){
                return false;
            }
        }

        for(int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--){
            if(cur[i][j] == 'Q'){
                return false;
            }
        }

        for(int i = row - 1, j = col + 1; i >= 0 && j < cur.size(); i--, j++){
            if(cur[i][j] == 'Q'){
                return false;
            }
        }

        return true;
    }

    void solveNQueensDFS(vector &cur, int row, int &cnt){
        if(row == cur.size()){
            cnt++;
            return;
        }

        for(int col = 0; col < cur.size(); col++){
            if(isValid(cur, row, col)){
                cur[row][col] = 'Q';
                solveNQueensDFS(cur, row + 1, cnt);
                cur[row][col] = '.';
            }
        }
    }
  
};

测试一下，

Success
Details
Runtime: 0 ms, faster than 100.00% of C++ online submissions for N-Queens II.
Memory Usage: 8.5 MB, less than 39.40% of C++ online submissions for N-Queens II.