问题：
具体的问题可以看：https://leetcode.com/problems/out-of-boundary-paths/
解答：
两种方法的思路其实是一样的，但是迭代比递归要快。
从计算量的角度看：
result_list[time][i][j] 表示，在i,j这个位置，能进行time次移动时的结果。
result_list[N][i][j] 即为所求结果。
迭代需要计算result_list[N][0~m-1][0~n-1]，而迭代只需要result_list[N][i][j]
我觉得计算量应该会更大，但是实际结果还是迭代比较快, 可能是递归本身的消耗比较大吧。
注意: python的深复制和浅复制
class Solution:
def findPaths(self, m, n, N, in_i, in_j):
"""
:type m: int
:type n: int
:type N: int
:type i: int
:type j: int
:rtype: int
"""
# 空间换时间
# result_list[time][i][j] 表示，在i,j这个位置，且能进行times移动时的结果
# result_list[time][i][j] = result_list[time-1][i-1][j] +
# result_list[time-1][i+1][j] +
# result_list[time-1][i][j-1] +
# result_list[time-1][i][j+1] //(如果出界，替换为1。)
#
# 创建存放的数组[time][i][j]
# 方法1：迭代形式
# 特殊情况
if N==0:
return 0
import copy
# 创建result_list
l_n = [0 for j in range(n)]
l_m = [copy.deepcopy(l_n) for i in range(m)]
result_list = [copy.deepcopy(l_m) for t in range(N+1)]
MOD = 1000000007
for time in range(1,N+1):
for i in range(m):
for j in range(n):
count = 0
if i-1>=0:
count = count+result_list[time-1][i-1][j]
else:
count = count+1
if i+1<m:
count = count+result_list[time-1][i+1][j]
else:
count = count+1
if j-1>=0:
count = count+result_list[time-1][i][j-1]
else:
count = count+1
if j+1<n:
count = count+result_list[time-1][i][j+1]
else:
count = count+1
result_list[time][i][j] = count % MOD
return result_list[N][in_i][in_j]
"""
# 方法2 递归 比较慢 超时
# 递归函数
def get_count(m, n, i ,j, time, result_list):
MOD = 1000000007
count = 0
if time==1:
#print("case1")
count = result_list[1][i][j]
else:
if result_list[time][i][j]!=0 :
#print("case2")
# 之前计算过，空间换时间
count = result_list[time][i][j]
else:
#print("case3")
#向上
if i-1>=0:
count = count + get_count(m, n, i-1 ,j, time-1, result_list)
else:
count = count + 1
#向下
if i+1<m:
count = count + get_count(m, n, i+1 ,j, time-1, result_list)
else:
count = count + 1
#向左
if j-1>=0:
count = count + get_count(m, n, i ,j-1, time-1, result_list)
else:
count = count + 1
#向右
if j+1<n:
count = count + get_count(m, n, i ,j+1, time-1, result_list)
else:
count = count + 1
result_list[time][i][j]=count%MOD
#print(i,j,time,count)
return result_list[time][i][j]
import copy
# 特殊情况
if N==0:
return 0
# 创建result_list
l_n = [0 for j in range(n)]
l_m = [copy.deepcopy(l_n) for i in range(m)]
result_list = [copy.deepcopy(l_m) for t in range(N+2)]
# 初始化
if m==1 and n==1:
result_list[1][0][0] = 4
if n==1 and m>1:
for i in range(1,m-1):
result_list[1][i][0] = 2
result_list[1][0][0] = result_list[1][m-1][0] = 3
if m==1 and n>1:
for j in range(1,n-1):
result_list[1][0][j] = 2
result_list[1][0][0] = result_list[1][0][n-1] = 3
if m>1 and n >1:
result_list[1][0][0] =2
result_list[1][0][n-1] =2
result_list[1][m-1][0] =2
result_list[1][m-1][n-1] =2
for i in range(1,m-1):
result_list[1][i][0] = result_list[1][i][n-1] =1
for j in range(1,n-1):
result_list[1][0][j] = result_list[1][m-1][j] =1
return get_count(m, n, in_i ,in_j, N, result_list)
"""
#=====================
s = Solution()
#print(s.findPaths(2,2,2,0,0)) #return 6
#print(s.findPaths(1,3,3,0,1)) #return 12
#print(s.findPaths(2,3,8,1,0)) #return 1104
#print(s.findPaths(3,2,5,1,1)) #return 109
#print(s.findPaths(3,2,5,1,1)) #return 109
#print(s.findPaths(30,24,23,26,12))