LeetCode51 N-Queens
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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 };
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