LeetCode51 N-Queens

题目：

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. （Hard）

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

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

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

分析：

n皇后问题，典型的搜索思路，对每一行，依次遍历选择一个位置，添加一个Q进去，判断是否合法。合法则处理下一行，不合法则回退到上一行选择其他位置添加Q。

注意isVaild函数的写法，行在添加过程中保证不重复，列需要判断，主副对角线通过x + y为定值和 x - y为定值判断（注意均只需要判断x，y之前的即添加过的部分）。

代码：

 1 class Solution {
 2 private:
 3     vector<vector<string>> result;
 4     void helper(vector<string>& v, int row, int n) {
 5         for (int i = 0; i < n; ++i) {
 6             v[row][i] = ‘Q‘;
 7             if (isValid(v, row, i, n)) {
 8                 if (row == n - 1) {
 9                     result.push_back(v);
10                     v[row][i] = ‘.‘;
11                     return;
12                 }
13                 helper(v, row + 1, n);
14             }
15             v[row][i] = ‘.‘;
16         }
17     }
18     bool isValid (const vector<string>& v, int x, int y, int n) {
19         for (int i = 0; i < x; ++i) {
20             if (v[i][y] == ‘Q‘) {
21                 return false;
22             }
23         }
24         for(int i = x - 1, j = y - 1; i >= 0 && j >= 0; i--,j--) {
25             if(v[i][j] == ‘Q‘) {
26                 return false;
27             }
28         }
29         for(int i = x - 1, j = y + 1; i >= 0 && j < n; i--,j++){
30             if(v[i][j] == ‘Q‘) {
31                 return false;
32             }
33         }
34         return true;
35     }
36 public:
37     vector<vector<string>> solveNQueens(int n) {
38         vector<string> v(n, string(n, ‘.‘));
39         helper(v,0,n);
40         return result;
41     }
42 };