[Leetcode] N-Queens 系列
N-Queens 系列题解
题目来源：
N-Queens
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
class Solution {
private:
int dimension;
inline bool onDiagonal(int r1, int r2, int c1, int c2) {
int rowDis = r1 - r2;
int colDis = c1 - c2;
return rowDis == colDis || -rowDis == colDis;
}
bool isValid(const vector<string>& board, int row, int col) {
if(row == 0)
return true;
for (int r = 0; r < row; r++) {
for (int c = 0; c < dimension; c++) {
if (board[r][c] == 'Q' &&
(c == col || onDiagonal(r, row, c, col)))
return false;
}
}
return true;
}
void backTrack(vector<vector<string>>& res,
vector<string>& board, int row) {
if (row == dimension) {
res.push_back(board);
} else {
for (int col = 0; col < dimension; col++) {
if (isValid(board, row, col)) {
board[row][col] = 'Q';
backTrack(res, board, row + 1);
board[row][col] = '.';
}
}
}
}
public:
vector<vector<string>> solveNQueens(int n) {
vector<vector<string>> res;
if (n == 0)
return res;
dimension = n;
vector<string> board(dimension, string(dimension, '.'));
backTrack(res, board, 0);
return res;
}
};
解题描述
这道题就是经典的N皇后问题，要求在给出的n×n的棋盘上放置n个皇后，皇后之前不会互相击杀（不在同一行、同一列、同一对角线上）。上面用到的是递归回溯的办法，每次确定新一行上的皇后，添加新一行的时候要检验与之前行上的皇后是否会相互击杀。
更优解法
2018.2.27更新
评估上面的解法，其实还是存在可以优化的空间。对整个递归回溯解法来说，对特定的维度n来说，递归的层数是固定的，所以时间复杂度主要取决于位置合法性检查，而细想可知，其实上面的解法中合法性检查部分做了很多不必要的操作：遍历棋盘来找到存在皇后的位置。而对整个棋盘而言，我们需要检查的位置其实对每个维度来说是固定的：
- n个列
- 2n - 1条左对角线
- 2n - 1条右对角线
权衡利弊，我们可以采用以空间换时间的办法，记录上述5n -2个位置上的皇后占据情况来降低检查的时间复杂度。下面给出具体实现：
class Solution {
private:
int dimension;
vector<bool> colFlags, leftDiaFlags, rightDiaFlags;
void backTrack(vector<vector<string>>& res,
vector<string>& board, int row) {
if (row == dimension) {
res.push_back(board);
return;
}
for (int col = 0; col < dimension; col++) {
if (!colFlags[col] &&
!leftDiaFlags[dimension - 1 + row - col] &&
!rightDiaFlags[row + col]) {
colFlags[col]
= leftDiaFlags[dimension - 1 + row - col]
= rightDiaFlags[row + col]
= true;
board[row][col] = 'Q';
backTrack(res, board, row + 1);
board[row][col] = '.';
colFlags[col]
= leftDiaFlags[dimension - 1 + row - col]
= rightDiaFlags[row + col]
= false;
}
}
}
public:
vector<vector<string>> solveNQueens(int n) {
vector<vector<string>> res;
if (n == 0)
return res;
dimension = n;
vector<string> board(dimension, string(dimension, '.'));
// 三种位置上的皇后是否存在的标记：
colFlags.resize(dimension, false); // 1. 列
leftDiaFlags.resize(2 * dimension - 1, false); // 2. 左对角线
rightDiaFlags.resize(2 * dimension - 1, false); // 3. 右对角线
backTrack(res, board, 0);
return res;
}
};
N-Queens II
这道题题意只要求求出N皇后解法数目，不要求给出具体解法，则与第一版差别不大，只需略作修改。
Solution
class Solution {
private:
int dimension;
vector<bool> colFlags, leftDiaFlags, rightDiaFlags;
void backTrack(int& res, int row) {
if (row == dimension) {
++res;
return;
}
for (int col = 0; col < dimension; col++) {
if (!colFlags[col] &&
!leftDiaFlags[dimension - 1 + row - col] &&
!rightDiaFlags[row + col]) {
colFlags[col]
= leftDiaFlags[dimension - 1 + row - col]
= rightDiaFlags[row + col]
= true;
backTrack(res, row + 1);
colFlags[col]
= leftDiaFlags[dimension - 1 + row - col]
= rightDiaFlags[row + col]
= false;
}
}
}
public:
int totalNQueens(int n) {
int res = 0;
if (n == 0)
return res;
dimension = n;
// 三种位置上的皇后是否存在的标记：
colFlags.resize(dimension, false); // 1. 列
leftDiaFlags.resize(2 * dimension - 1, false); // 2. 左对角线
rightDiaFlags.resize(2 * dimension - 1, false); // 3. 右对角线
backTrack(res, 0);
return res;
}
};
