1 """ 2 A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). 3 The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). 4 How many possible unique paths are there? 5 Above is a 7 x 3 grid. How many possible unique paths are there? 6 Note: m and n will be at most 100. 7 Example 1: 8 Input: m = 3, n = 2 9 Output: 3 10 Explanation: 11 From the top-left corner, there are a total of 3 ways to reach the bottom-right corner: 12 1. Right -> Right -> Down 13 2. Right -> Down -> Right 14 3. Down -> Right -> Right 15 Example 2: 16 Input: m = 7, n = 3 17 Output: 28 18 """ 19 """ 20 看见这个题，我找了个规律，用了递归 21 但对于数据量大时 会超时 22 Time Limit Exceeded 23 Last executed input 24 23 25 12 26 ------------------------------------------------- 27 以下为手动计算 28 1 1 1 1 29 1 2 3 4 30 1 3 6 10 31 paths[i, j] == paths[i-1, j] + paths[i, j-1] 32 """ 33 class Solution1: 34 def uniquePaths(self, m, n): 35 if m == 1 or n == 1: 36 return 1 37 else: 38 return self.uniquePaths(m-1, n)+self.uniquePaths(m, n-1) 39 40 """ 41 解法二可以用空间换时间，建立一个二维数组 42 数组的[-1,-1]就是路径数 43 """ 44 class Solution2: 45 def uniquePaths(self, m, n): 46 dp = [[1]*m for _ in range(n)] #n行 m列的初始化为1的矩阵 47 for i in range(1, n): 48 for j in range(1, m): 49 dp[i][j] = dp[i-1][j] + dp[i][j-1] 50 return dp[-1][-1] 51 52 """ 53 解法三用了一定数学基础 54 设向下走是1，向右走是0， 55 原问题变成m-1个0与n-1个1有多少种不同的排列，高中数学 56 思考在 （m-1）+（n-1）个位置里放m-1个0，其余n-1个位置自动为1 57 就是(m+n-2)!//((m-1)!*(n-1)!) 58 """ 59 class Solution3: 60 def uniquePaths(self, m, n): 61 t = 1 62 for i in range(1, n): #！！！bug因为(1，n)实际就是(n-1)! 63 t = t * i 64 e = 1 65 for i in range(m, m+n-1):#这里是(m+n-2)!//(m-1)! 66 e = e * i 67 return e // t #这里即为(m+n-2)!//((m-1)!*(n-1)!)