[LeetCode] 62. Unique Paths（不同的路径）

• Difficulty: Medium

• Related Topics: Array, Dynamic Programming

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

Description

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
一只机器人位于 m x n 方格的左上角（下图中以“Start”标识）。

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).
机器人只能向下或向右移动。该机器人欲抵达方格的右下角（下图中以“Finish”标识）。

How many possible unique paths are there?
有多少种不同的路径？

Examples

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

Example 3

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

Example 4

Input: m = 3, n = 3
Output: 6

Constraints

• 1 <= m, n <= 100

• It's guarranteed that the answer will be less than or equal to 2 * 10^9.

Solution

动态规划例题之二，状态转义方程如下：

\[\texttt{dp(i, j)} = \begin{cases} 1, &i = 0 或j = 0\\ \texttt{dp(i - 1, j)} + \texttt{dp(i, j - 1)}, &else \end{cases} \]

其中 dp(i, j) 表示 (i, j) 时可能的路径数

class Solution {
    fun uniquePaths(m: Int, n: Int): Int {
        val dp = Array(m) { IntArray(n) }
        (0 until m).forEach { dp[it][0] = 1 }
        (0 until n).forEach { dp[0][it] = 1 }

        for (i in 1 until m) {
            for (j in 1 until n) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
            }
        }

        return dp.last().last()
    }
}