# 62. Unique Paths
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?

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 = 3, n = 3
Output: 6

Constraints:

1 <= m, n <= 100

It’s guaranteed that the answer will be less than or equal to 2 * 109.

1. 动态规划解法

class Solution {public int uniquePaths(int m, int n) {int[][] dp = new int[m][n];for (int i = 0; i < m; i++) dp[i][0] = 1;for (int i = 0; i < n; i++) dp[0][i] = 1;for (int r = 1; r < m; r++) {for (int c = 1; c < n; c++) {dp[r][c] = dp[r - 1][c] + dp[r][c - 1];}}return dp[m - 1][n - 1];}
}

2. 排列组合解法

这是一个组合问题，可以在没有 DP 的情况下解决。对于 mxn 网格，机器人必须准确地向下移动 m-1 步和向右移动 n-1 步，这些可以以任何顺序完成。

例如，给出的问题是 3x7 矩阵，机器人需要以任意顺序执行 2+6 = 8 步，其中 2 步向下，6 步向右。那不过是一个排列问题。向下表示为“D”，向右表示为“R”，以下是路径之一：-

DRRRDRRR

我们必须告诉上面给定单词的排列总数。因此，将 m & n 都减少 1 并应用以下公式：-

总排列 = (m+n)！/（m！* n！）

以下是我的代码做同样的事情：-

class Solution {public int uniquePaths(int m, int n) {if (m == 1 || n == 1) return 1;m--;n--;int hi = m > n ? m : n;int lo = m > n ? n : m;long r = 1;for (int i = 1, k = hi + 1; k <= hi + lo; i++, k++) {r = r * k / i;}return (int)r;}
}

参考

https://leetcode.com/problems/unique-paths/discuss/22953/Java-DP-solution-with-complexity-O(n*m)

https://leetcode.com/problems/unique-paths/discuss/22958/Math-solution-O(1)-space

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算法: 唯一路径62. Unique Paths

1. 动态规划解法

2. 排列组合解法

参考

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