Knight Probability in Chessboard

https://leetcode.com/problems/knight-probability-in-chessboard/description/

On an NxN chessboard, a knight starts at the r-th row and c-th column and attempts to make exactly K moves. The rows and columns are 0 indexed, so the top-left square is (0, 0), and the bottom-right square is (N-1, N-1).

A chess knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.

Each time the knight is to move, it chooses one of eight possible moves uniformly at random (even if the piece would go off the chessboard) and moves there.

The knight continues moving until it has made exactly K moves or has moved off the chessboard. Return the probability that the knight remains on the board after it has stopped moving.

Example:

Input: 3, 2, 0, 0

Output: 0.0625

Explanation: There are two moves (to (1,2), (2,1)) that will keep the knight on the board.

From each of those positions, there are also two moves that will keep the knight on the board.

The total probability the knight stays on the board is 0.0625.

Thoughts

给定Ｎ＊Ｎ矩阵，设F(i, j, k)为走k步还在格子上的概率.

先来看看例子是怎么算的，1/8 * 2/8 + 1/8*2/8 = 0.0625，其中1/8为从当前位置（0,0）移动一步（第二步）能到达的位置（1，2）的概率，2/8是从(1,2)移动一步（第一步，也就是最后一步）还在棋盘的概率，也就是说从我们当前位置移动K步还在棋盘的概率为sum_xy(P_x,y(K-1)). 当K为0，则f(i,j)肯定在棋盘上所以概率为1.

Code

class Solution {
    public double knightProbability(int N, int K, int r, int c) {
        double[][][] f = new double[N][N][K + 1];

        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                f[i][j][0] = 1.0;
            }
        }

         for (int k = 1; k <= K; k++) {                                    
            for (int i = 0; i < N; i++) {
                for (int j = 0; j < N; j++) {
                    if (i - 1 >= 0) {
                        if (j - 2 >= 0) {
                            f[i][j][k] += 0.125 * f[i - 1][j - 2][k - 1];
                        }
                        if (j + 2 < N) {
                            f[i][j][k] += 0.125 * f[i - 1][j + 2][k - 1];
                        }
                        if (i - 2 >= 0) {
                            if (j - 1 >= 0) {
                                f[i][j][k] += 0.125 * f[i - 2][j - 1][k - 1];
                            }
                            if (j + 1 < N) {
                                f[i][j][k] += 0.125 * f[i - 2][j + 1][k - 1];
                            }
                        }                                                  
                    }
                    if (i + 1 < N) {
                        if (j - 2 >= 0) {
                            f[i][j][k] += 0.125 * f[i + 1][j - 2][k - 1];
                        }
                        if (j + 2 < N) {
                            f[i][j][k] += 0.125 * f[i + 1][j + 2][k - 1];
                        }
                        if (i + 2 < N) {
                            if (j - 1 >= 0) {
                                f[i][j][k] += 0.125 * f[i + 2][j - 1][k - 1];
                            }
                            if (j + 1 < N) {
                                f[i][j][k] += 0.125 * f[i + 2][j + 1][k - 1];
                            }
                        }
                    }
                }
            }
         }

        return f[r][c][K];
    }
}

Analysis

时间复杂度为O(mnk)

Ver.2

把能走的相对位移提前存下来，能减少代码量。

class Solution {
    public double knightProbability(int N, int K, int r, int c) {
        double[][][] f = new double[N][N][K + 1];
        int[][] ts = new int[][]{{1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}, {-2, 1}, {-1, 2}};

        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                f[i][j][0] = 1.0;
            }
        }

         for (int k = 1; k <= K; k++) {                                    
            for (int i = 0; i < N; i++) {
                for (int j = 0; j < N; j++) {
                    for (int[]t : ts) {
                        int x = i + t[0];
                        int y = j + t[1];
                        if (x < 0 || x >= N || y < 0 || y >= N) {
                            continue;
                        }
                        f[i][j][k] += 0.125 * f[x][y][k - 1];
                    }
                }
            }
         }

        return f[r][c][K];
    }
}

节约空间版, 别忘了重置.

class Solution {
    public double knightProbability(int N, int K, int r, int c) {
        int[][] dirs = new int[][]{{-2, -1}, {-1, -2}, {1, -2}, {2, -1}, {2, 1}, {-1, 2}, {-2, 1}, {1, 2}};
        double[][][] f = new double[N][N][2];

        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                f[i][j][0] = 1;
            }
        }

        for (int k = 1; k <= K; k++) {
            for (int i = 0; i < N; i++) {
                for (int j = 0; j < N; j++) {
                    f[i][j][k % 2] = 0;
                    for (int[] dir : dirs) {
                        int x = i + dir[0];
                        int y = j + dir[1];
                        if (x >= 0 && x < N && y >= 0 && y < N) {
                            f[i][j][k % 2] += 0.125 * f[x][y][(k - 1) % 2];
                        }
                    }
                }
            }
        }

        return f[r][c][K % 2];
    }
}