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.."] ]
解N皇后问题。在nXn的矩阵上放置n个皇后，每行放一个，其所在列和两个斜线上没有其他皇后放置。
该题要求得到所有可行解。‘Q’表示皇后，‘.‘表示空位置
class Solution {
public:
bool isValid(vector<string>&matrix, int x, int y, int n){
//考虑纵向
int i=0;
int j=y;
for(;i<n; i++){
if(i!=x && matrix[i][j]=='Q')return false;
}
//考虑指向左上角的斜线
i=x-1;
j=y-1;
while(i>=0&&j>=0){
if(matrix[i][j]=='Q')return false;
i--;j--;
}
//考虑指向右上角的斜线
i=x-1;
j=y+1;
while(i>=0&&j<n){
if(matrix[i][j]=='Q')return false;
i--;j++;
}
return true;
}
void dfs(vector<vector<string> >&result, vector<string>matrix, int lineNo, int n){
//matrix--已经完成部分皇后放置的二维矩阵
//lineNo--当前正在处理的行的行号，从1开始
//n--总行数，也就是皇后数
if(lineNo>n){
result.push_back(matrix);
return;
}
for(int i=0; i<n; i++){
matrix[lineNo-1][i]='Q';
if(isValid(matrix, lineNo-1, i, n)){
dfs(result, matrix, lineNo+1, n);
}
matrix[lineNo-1][i]='.';
}
}
vector<vector<string> > solveNQueens(int n) {
vector<vector<string> >result;
vector<string>matrix(n, string(n, '.'));
dfs(result, matrix, 1, n);
return result;
}
};LeetCode: N-Queens [050],布布扣,bubuko.com
原文地址：http://blog.csdn.net/harryhuang1990/article/details/26687153
