题目描述：
LeetCode 688. Knight Probability in Chessboard
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:
Nwill be between 1 and 25.Kwill be between 0 and 100.- The knight always initially starts on the board.
题目大意：
给定NxN棋盘，骑士初始位于r行c列。骑士每次等概率的向8个方向移动。
求骑士行进K步，落在棋盘内的概率。
解题思路：
动态规划（Dynamic Programming）
利用字典dmap存储骑士落在棋盘某位置的概率。
dmap[(nx, ny)] += dmap[(x, y)] * 0.125
其中(x, y)为当前位置，(nx, ny)为下一个位置
Python代码：
class Solution(object):
def knightProbability(self, N, K, r, c):
"""
:type N: int
:type K: int
:type r: int
:type c: int
:rtype: float
"""
dxs = [-2, -2, -1, -1, 1, 1, 2, 2]
dys = [-1, 1, -2, 2, -2, 2, -1, 1]
ans = 0
dmap = {(r, c) : 1}
for t in range(K):
dmap0 = collections.defaultdict(int)
for (x, y), pb in dmap.iteritems():
for dx, dy in zip(dxs, dys):
nx, ny = x + dx, y + dy
if nx < 0 or nx >= N or ny < 0 or ny >= N:
ans += 0.125 * pb
else:
dmap0[(nx, ny)] += 0.125 * pb
dmap = dmap0
return 1 - ans
本文链接：http://bookshadow.com/weblog/2017/10/01/leetcode-knight-probability-in-chessboard/
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