Leetcode 62. Unique Paths

https://leetcode.com/problems/unique-paths/

Medium

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:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

Input: m = 7, n = 3
Output: 28

---

• 動歸。 dp[i][j]表明[i,j]的unique path，dp[i][j] = dp[i - 1][j] + dp[i][j - 1]，dp[0, 0] = 1。
• 第一種解法最簡單直接。
• 分析第一種解法代碼發現，實際上只須要保存previous row和current row結果，由此得出第二種解法。
• 再分析第二種解法代碼發現，previous[j]實際上就是更新前的current[j]，由此只須要保存current row結果就行。
• https://leetcode.com/problems/unique-paths/discuss/22954/C%2B%2B-DP

 1 class Solution:
 2     def uniquePaths1(self, m: int, n: int) -> int:
 3         dp = [ [ 1 for x in range(n) ] for x in range(m) ]
 4         
 5         for i in range(1, m):
 6             for j in range(1, n):
 7                 dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
 8         
 9         return dp[m - 1][n - 1]        
10 
11     def uniquePaths2(self, m: int, n: int) -> int:
12         previous, current = [1] * n, [1] * n
13         
14         for i in range(1, m):
15             for j in range(1, n):
16                 current[j] = previous[j] + current[j - 1]
17             
18             current, previous = previous, current
19         
20         # note : previous not current
21         return previous[n - 1]
22     
23     def uniquePaths(self, m: int, n: int) -> int:
24         current = [1] * n
25         
26         for i in range(1, m):
27             for j in range(1, n):
28                 current[j] += current[j - 1]
29         
30         return current[n - 1]

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