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62.unique-paths

Statement

Metadata
  • Link: 不同路径
  • Difficulty: Medium
  • Tag: 数学 动态规划 组合数学

一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为 “Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为 “Finish” ）。

问总共有多少条不同的路径？

 

示例 1：

输入：m = 3, n = 7
输出：28

示例 2：

输入：m = 3, n = 2
输出：3
解释：
从左上角开始，总共有 3 条路径可以到达右下角。
1. 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右
3. 向下 -> 向右 -> 向下

示例 3：

输入：m = 7, n = 3
输出：28

示例 4：

输入：m = 3, n = 3
输出：6

 

提示：

  • 1 <= m, n <= 100
  • 题目数据保证答案小于等于 2 * 109

Metadata
  • Link: Unique Paths
  • Difficulty: Medium
  • Tag: Math Dynamic Programming Combinatorics

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 * 109.

 

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

Solution

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        f = [[0 for i in range(n + 1)] for j in range(m + 1)]
        f[0][0] = 1
        for i in range(m):
            for j in range(n):
                if i - 1 >= 0:
                    f[i][j] += f[i - 1][j]
                if j - 1 >= 0:
                    f[i][j] += f[i][j - 1]
        return f[m - 1][n - 1]

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function (s, p) {
    const r = new RegExp(`^${p}$`);
    return r.test(s);
};
最后更新: October 11, 2023
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