[Leetcode]62. Unique Paths
62. Unique Paths
- 本题难度: Easy
- Topic: Dynamic Programming
Description
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
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).
How many possible unique paths are there?
Above is a 7 x 3 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
Example 1:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
- Right -> Right -> Down
- Right -> Down -> Right
- Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
我的代码
recursion
class Solution:
def uniquePaths(self, m: 'int', n: 'int') -> 'int':
if m < 0 or n<0:
return 0
if m==1 and n==1:
return 1
return self.uniquePaths(m-1,n)+self.uniquePaths(m,n-1)
no recursion
class Solution:
def uniquePaths(self, m: 'int', n: 'int') -> 'int':
res = [[0 for _ in range(m+1)] for _ in range(n+1)]
res[1][1] = 1
for i in range(1,n+1):
for j in range(1,m+1):
if j==1 and i==1:
continue
res[i][j] = res[i-1][j]+res[i][j-1]
return res[n][m]
思路
recurison
到点(m,n)有f(m,n)中走法
每一个点都可以从它的左边或者上面到达，所以f(m,n-1) +f(m,n-1)
但是时间复杂度超过了，所以我用了一个二维数组存储之前的值。
- 时间复杂度
recursion
时间复杂度 O((mn)^2) 空间复杂度O(1)
no recursion
时间复杂度 O(mn) 空间复杂度O(mn) - 出错
