LeetCode Knight Probability in Chessboard

Posted 2020-10-09 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/knight-probability-in-chessboard/description/

题目：

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.

 

Note:

• N will be between 1 and 25.
• K will be between 0 and 100.
• The knight always initially starts on the board.

题解：
类似Out of Boundary Paths.

DP问题. 求最后在board上的概率. 反过来想，走完K步棋子在board上的哪个位置呢. 反过来走, 看board上所有位置走完K步后能到初始位置(r,c)的数目和.

储存历史信息是走到当前这步时棋盘上能走到每个位置的不同走法.

递推时, 向所有方向移动, 若是还在board上就把自己的走法加到新位置的走法上.

初始化所有位置只有1种走法.

答案K步之后到初始位置的走法除以Math.pow(8,K).

Time Complexity: O(K*N^2).

Space: O(N^2).

AC Java:

 1 class Solution {
 2     public double knightProbability(int N, int K, int r, int c) {
 3         int [][] moves = {{1,2},{1,-2},{2,1},{2,-1},{-1,2},{-1,-2},{-2,1},{-2,-1}};
 4         double [][] dp0 = new double[N][N];
 5         for(double [] row : dp0){
 6             Arrays.fill(row, 1);
 7         }
 8         
 9         for(int step = 0; step<K; step++){
10             double [][] dp1 = new double[N][N];
11             for(int i = 0; i<N; i++){
12                 for(int j = 0; j<N; j++){
13                     for(int [] move : moves){
14                         int row = i + move[0];
15                         int col = j + move[1];
16                         if(isIllegal(row, col, N)){
17                             dp1[row][col] += dp0[i][j];
18                         }
19                     }
20                 }
21             }
22             dp0 = dp1;
23         }
24         return dp0[r][c]/Math.pow(8,K);
25     }
26     
27     private boolean isIllegal(int row, int col, int len){
28         return row>=0 && row<len && col>=0 && col<len;
29     }
30 }