LeetCode 062 Unique Paths

LeetCode

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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:

<pre>Input: m = 3, n = 7 Output: 28 </pre>

Example 2:

<pre>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 </pre>

Constraints:

• 1 <= m, n <= 100

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题目大意：

求矩阵路径总数

解题思路：

求个数用DP，递归式：

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

解题步骤：

N/A

注意事项：

1. 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
2. range(m)不是range(len(m))
3. 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<code>O(n²)</code>，空间复杂度<code>O(n²)</code>