LeetCode 062 Unique Paths

LeetCode

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to 2 * 10<sup>9</sup>.

Example 1:

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

Example 2:

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 -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down

Constraints:

* 1 <= m, n <= 100

题目大意：

求矩阵路径总数

解题思路：

求个数用DP，递归式：

1	dp[i][j] = dp[i-1][j] + dp[i][j-1]

解题步骤：

N/A

注意事项：

初始值dp[1] = 1而不是dp[0] = 1因为第二行的第一格不能加左边的虚拟格=1
range(m)不是range(len(m))
优化空间用一维

Python代码：

def uniquePaths(self, m: int, n: int) -> int:
	dp = [0] * (n + 1)
	dp[1] = 1 # remember not dp[0] = 1
	for i in range(m): # remember no len(m)
		for j in range(1, len(dp)):
			dp[j] += dp[j - 1]
	return dp[-1]

算法分析：

时间复杂度为O(n²)，空间复杂度O(n²)