Problem
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?
Above is a 7 x 3 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
Example 1:
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 -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
Solution
/**
* @param {number} m
* @param {number} n
* @return {number}
*/
var uniquePaths = function(m, n) {
var dp = Array(m);
if (!m || !n) return 0;
for (var i = 0; i < m; i++) {
dp[i] = 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];
};
Explain:
dp[i][j] 代表到达该点的路径数量。该点可以从左边点或上边点到达
也就是 dp[i][j] = dp[i - 1][j] + dp[i][j - 1] 。
Complexity:
- Time complexity : O(m*n).
- Space complexity : O(m*n).