N-Queens

2014.2.13 19:23

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 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.."]
]

Solution:

  The Eight Queens problem is a typical model for backtracking  algorithm.

  For any pair of queens, their difference in x and y coordinates mustn't be 0 or equal, that's on the same row, column or diagonal line.

  The code is short and self-explanatory, please see for yourself.

  Total time complexity is O(n!). Space complexity is O(n!) as well, which comes from local parameters in recursive function calls.

Accepted code:

 1 // 1CE, 1WA, 1AC, try to make the code shorter, it'll help you understand it better.
 2 class Solution {
 3 public:
 4     vector<vector<string> > solveNQueens(int n) {
 5         a = nullptr;
 6         res.clear();
 7         if (n <= 0) {
 8             return res;
 9         }
10         
11         a = new int[n];
12         solveNQueensRecursive(0, a, n);
13         delete[] a;
14         
15         return res;
16     }
17 private:
18     int *a;
19     vector<vector<string> > res;
20     
21     void solveNQueensRecursive(int idx, int a[], const int &n) {
22         if (idx == n) {
23             // one solution is found
24             addSingleResult(a, n);
25             return;
26         }
27         
28         int i, j;
29         // check if the current layout is valid.
30         for (i = 0; i < n; ++i) {
31             a[idx] = i;
32             for (j = 0; j < idx; ++j) {
33                 if (a[j] == a[idx] || myabs(idx - j) == myabs(a[idx] - a[j])) {
34                     break;
35                 }
36             }
37             if (j == idx) {
38                 // valid layout.
39                 solveNQueensRecursive(idx + 1, a, n);
40             }
41         }
42     }
43     
44     void addSingleResult(const int a[], int n) {
45         vector<string> single_res;
46         char *str = nullptr;
47         
48         str = new char[n + 1];
49         int i, j;
50         for (i = 0; i < n; ++i) {
51             for (j = 0; j < n; ++j) {
52                 str[j] = '.';
53             }
54             str[j] = 0;
55             str[a[i]] = 'Q';
56             single_res.push_back(string(str));
57         }
58         
59         res.push_back(single_res);
60         single_res.clear();
61         delete []str;
62     }
63     
64     int myabs(const int x) {
65         return (x >= 0 ? x : -x);
66     }
67 };

 

 posted on 2014-02-13 19:45  zhuli19901106  阅读(344)  评论(0编辑  收藏  举报