力扣面试题 08.12. 八皇后(java回溯解法)
Problem: 面试题 08.12. 八皇后
题目描述
思路
八皇后问题的性质可以利用回溯来解决,将大问题具体分解成如下待解决问题:
1.以棋盘的每一行为回溯的决策阶段,判断当前棋盘位置能否放置棋子
2.如何判断当前棋盘位置是否可以放置棋子
解题方法
1.回溯函数:
1.1定义二维结果集result,char类型二维数组(作为棋盘)并初始化
1.2当决策阶段row等于n时,将当前的决策路径添加到result中(注意决策阶段应该等于n时才说明将棋盘判断完了,因为当决策阶段等于n时说明0 - n-1 已经判断处理完)
1.3由于在每一个决策阶段我们需要对棋盘的每一列棋格判断(穷举),所以以每一列为循环判断(调用判断当前位置是否可以添加棋子的函数),若可以则先将棋盘当前位置添上棋子,再回溯判断当前行的下一行,判断完当前行后还需恢复当前棋盘位置的状态
2.判断当前位置是否可以添加棋子函数
2.1依次利用循环判断当前位置的列,右上角,左上角是否存在棋子,存在则不可在当前位置添加棋子
复杂度
时间复杂度:
O ( n ! ) O(n!) O(n!)
空间复杂度:
O ( n ) O(n) O(n)
Code
class Solution {
//Two-dimensional result set
private List<List<String>> result = new ArrayList<>();
/**
* Get the solution to the Eight Queens problem
*
* @param n The size of board
* @return List<List < String>>
*/
public List<List<String>> solveNQueens(int n) {
char[][] board = new char[n][n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
board[i][j] = '.';
}
}
backtrack(0, board, n);
return result;
}
/**
* Find the solution of the eight queens problem by backtracking
*
* @param board Board
* @param row The row of board(The decision stage of backtracking)
* @param n The size of board
*/
private void backtrack(int row, char[][] board, int n) {
//End condition:A feasible solution is found
if (row == n) {
List<String> snapshot = new ArrayList<>();
for (int i = 0; i < n; ++i) {
snapshot.add(new String(board[i]));
}
result.add(snapshot);
return;
}
//Each has n ways to place
for (int col = 0; col < n; ++col) {
if (isOk(board, n, row, col)) {//optional list
//The chess board places pieces in row rows and col columns
board[row][col] = 'Q';
//Investigate the next row
backtrack(row + 1, board, n);
//Recover the selection
board[row][col] = '.';
}
}
}
/**
* Determines whether the current column can place chess pieces
*
* @param board The board
* @param n The row number and column number of board
* @param row The row number of board
* @param col The column number of board
* @return boolean
*/
private boolean isOk(char[][] board, int n, int row, int col) {
//Check whether columns conflict
for (int i = 0; i < n; ++i) {
if (board[i][col] == 'Q') {
return false;
}
}
//Check whether top right corner conflict
int i = row - 1;
int j = col + 1;
while (i >= 0 && j < n) {
if (board[i][j] == 'Q') {
return false;
}
i--;
j++;
}
//Check whether top left corner conflict
i = row - 1;
j = col - 1;
while (i >= 0 && j >= 0) {
if (board[i][j] == 'Q') {
return false;
}
i--;
j--;
}
return true;
}
}
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