题解 | #动态规划# #不同路径的数目(一)#

#coding:utf-8
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param m int整型 
# @param n int整型 
# @return int整型
#
class Solution:
    def uniquePaths(self , m , n ):
        # write code here
        ##init
        if m <= 0:
            return 0
        if n <= 0:
            return 0
        
        dp = [[0] * n for i in range(0, m)]
        #print ("dp: ", dp)
        
        ##process
        ##dp[i][j]：到达i，j位置的路径数
        ##transition: dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        ##corner: dp[0][j] = 1, dp[i][0] = 1
        for i in range(0, m):
            dp[i][0] = 1
        for j in range(0, n):
            dp[0][j] = 1
        
        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        #print ("New dp: ", dp)

        return dp[m - 1][n - 1]