688. Knight Probability in Chessboard

688. Knight Probability in Chessboard

• 题目描述: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, 0Output: 0.0625Explanation: 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.

• 题目大意：给定一个N*N的棋盘，骑士每次只能在8个方向随机的选择一个方向移动，问经过K步之后，骑士是否还在棋盘内。

• 代码

  package DP;/*** @Author OovEver* @Date 2017/12/16 14:29*/public class LeetCode688 {  int[][] moves = {{1, 2}, {1, -2}, {2, 1}, {2, -1}, {-1, 2}, {-1, -2}, {-2, 1}, {-2, -1}};  public double knightProbability(int N, int K, int r, int c) {      double[][][] dp = new double[K + 1][N][N];      return helper(dp, N, K, r, c) / Math.pow(8.0, K);  }  private double helper(double[][][] dp, int N, int k, int r, int c) {      if(r<0||r>=N||c<0||c>=N) return 0.0;      if (k == 0) {          return 1.0;      }//        如果遇到之前遍历过的直接返回      if (dp[k][r][c] != 0.0) return dp[k][r][c];      for(int i=0;i<8;i++) {          dp[k][r][c] += helper(dp, N, k - 1, r + moves[i][0], c + moves[i][1]);      }      return dp[k][r][c];  }}