深夜十大禁用色多多seo网站建设是什么意思

目录

题目描述

方法一、回溯+每次判断是否合法

方法二、回溯+哈希

方法三、回溯+位运算

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题目描述

51. N-Queens

方法一、回溯+每次判断是否合法

class Solution {vector<vector<string>> res;vector<string> chessboard;
public:vector<vector<string>> solveNQueens(int n) {chessboard.resize(n,string(n,'.'));backtrack(n,0);return res;}void backtrack(int n,int row){if(row == n){res.push_back(chessboard);return;}for(int col = 0;col < n;col++){if(valid(row,col,n)){chessboard[row][col] = 'Q';backtrack(n,row+1);chessboard[row][col] = '.';}}}bool valid(int row,int col,int n){for(int i = row-1;i>=0;i--){if(chessboard[i][col] == 'Q')return false;}for(int i = row-1,j = col-1;i>=0&&j>=0;i--,j--){if(chessboard[i][j] == 'Q')return false;}for(int i = row-1,j = col+1;i>=0&&j<n;i--,j++){if(chessboard[i][j] == 'Q')return false;}return true;}
};

方法二、回溯+哈希

记录棋盘上已经处于攻击范围的直线，将判断能否摆放新皇后的时间复杂度降为O（1）。

class Solution {vector<vector<string>> res;vector<string> chessboard;vector<bool> columns;vector<bool> diagonals1;vector<bool> diagonals2;
public:vector<vector<string>> solveNQueens(int n) {chessboard.resize(n,string(n,'.'));columns.resize(n,false);//分别有2n-1条正斜线和反斜线diagonals1.resize(2*n-1,false);diagonals2.resize(2*n-1,false);backtrack(n,0);return res;}void backtrack(int n,int row){if(row == n){res.push_back(chessboard);return;}for(int col = 0;col < n;col++){if(columns[col])continue;int diag1 = row+col;if(diagonals1[diag1])continue;int diag2 = row-col+(n-1);if(diagonals2[diag2])continue;columns[col] = true;diagonals1[diag1] = true;diagonals2[diag2] = true;chessboard[row][col] = 'Q';backtrack(n,row+1);chessboard[row][col] = '.';columns[col] = false;diagonals1[diag1] = false;diagonals2[diag2] = false;}}
};

方法三、回溯+位运算

方法二中的哈希表可以改用位运算，节省空间开销。

class Solution {vector<vector<string>> res;vector<string> chessboard;int columns;   //第i个比特位为0表示第i列能放，为1表示不能放int diagonals1;//第i个比特位为0表示第i条反斜线能放，为1表示不能放int diagonals2;//正斜线的情况
public:vector<vector<string>> solveNQueens(int n) {chessboard.resize(n,string(n,'.'));columns = 0;//分别有2n-1条正斜线和反斜线,题目保证n<=9,那么2n-1<=17,int类型一共有32个比特位够用diagonals1 = 0;diagonals2 = 0;backtrack(n,0);return res;}void backtrack(int n,int row){if(row == n){res.push_back(chessboard);return;}for(int col = 0;col < n;col++){if(columns &(1<<col))continue;int diag1 = row+col;if(diagonals1 &(1<<diag1))continue;int diag2 = row-col+(n-1);if(diagonals2 &(1<<diag2))continue;columns    |= (1<< col);diagonals1 |= (1<< diag1);diagonals2 |= (1<< diag2);chessboard[row][col] = 'Q';backtrack(n,row+1);chessboard[row][col] = '.';columns    &= (~(1<< col));diagonals1 &= (~(1<<diag1));diagonals2 &= (~(1<<diag2));}}
};