knight-probability-in-chessboard

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Algorithms

On an <code>n x n</code> chessboard, a knight starts at the cell <code>(row, column)</code> and attempts to make exactly <code>k</code> moves. The rows and columns are 0-indexed, so the top-left cell is <code>(0, 0)</code>, and the bottom-right cell is <code>(n - 1, n - 1)</code>.

A chess knight has eight possible moves it can make, as illustrated below. Each move is two cells in a cardinal direction, then one cell in an orthogonal direction.
<img src="https://assets.leetcode.com/uploads/2018/10/12/knight.png" style="width: 300px; height: 300px;" />
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 <code>k</code> moves or has moved off the chessboard.

Return the probability that the knight remains on the board after it has stopped moving.

&nbsp;
Example 1:

<pre>
Input: n = 3, k = 2, row = 0, column = 0
Output: 0.06250
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.
</pre>

Example 2:

<pre>
Input: n = 1, k = 0, row = 0, column = 0
Output: 1.00000
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= n &lt;= 25</code></li>
	<li><code>0 &lt;= k &lt;= 100</code></li>
	<li><code>0 &lt;= row, column &lt;= n - 1</code></li>
</ul>

knightProbability

Knight Probability in Chessboard

Out of Boundary Paths

out-of-boundary-paths

出界的路径数

Maximum Number of Moves to Kill All Pawns

maximum-number-of-moves-to-kill-all-pawns

吃掉所有兵需要的最多移动次数

Dynamic Programming

在一个&nbsp;<code>n x n</code>&nbsp;的国际象棋棋盘上，一个骑士从单元格 <code>(row, column)</code>&nbsp;开始，并尝试进行 <code>k</code> 次移动。行和列是 从 0 开始 的，所以左上单元格是 <code>(0,0)</code> ，右下单元格是 <code>(n - 1, n - 1)</code> 。

象棋骑士有8种可能的走法，如下图所示。每次移动在基本方向上是两个单元格，然后在正交方向上是一个单元格。

<img src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/12/knight.png" style="height: 300px; width: 300px;" />

每次骑士要移动时，它都会随机从8种可能的移动中选择一种(即使棋子会离开棋盘)，然后移动到那里。

骑士继续移动，直到它走了 <code>k</code> 步或离开了棋盘。

返回 骑士在棋盘停止移动后仍留在棋盘上的概率 。

&nbsp;

示例 1：

<pre>
输入: n = 3, k = 2, row = 0, column = 0
输出: 0.0625
解释: 有两步(到(1,2)，(2,1))可以让骑士留在棋盘上。
在每一个位置上，也有两种移动可以让骑士留在棋盘上。
骑士留在棋盘上的总概率是0.0625。
</pre>

示例 2：

<pre>
输入: n = 1, k = 0, row = 0, column = 0
输出: 1.00000
</pre>

&nbsp;

提示:

<ul>
	<li><code>1 &lt;= n &lt;= 25</code></li>
	<li><code>0 &lt;= k &lt;= 100</code></li>
	<li><code>0 &lt;= row, column &lt;= n - 1</code></li>
</ul>