  [LeetCode][Medium] 62. Unique Paths
2021-04-03
Engineering
2 MIN

## 題目

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 = 7, n = 3
Output: 28
``````

Example 4:

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

## 思路

dp[i][j] 代表到達該點的路徑數量。而該點可以從左邊或上面到達 也就是 dp[i][j] = dp[i - 1][j] + dp[i][j - 1] 。

## Solution

``````/**
* @param {number} m
* @param {number} n
* @return {number}
*/
var uniquePaths = function(m, n) {
var dp = new Array(m)

if (!m || !n) return 0

for (var i = 0; i < m; i++) {
dp[i] = new Array(n)
for (var j = 0; j < n; j++) {
if (j > 0 && i > 0) dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
else if (j > 0 && i === 0) dp[i][j] = dp[i][j - 1]
else if (j === 0 && i > 0) dp[i][j] = dp[i - 1][j]
else dp[i][j] = 1
}
}
return dp[m - 1][n - 1]
};
``````

## Complexity

• Time complexity : O(m * n)
• Space complexity : O(m * n)