Question:

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

Thinking:

  • Method:
    • 对于第一列和第一行，都只可能为1,。
    • 对于其余的位置，为左侧和上侧的和。因为根据运动规则，只可能是从左侧和上侧来到该位置。
class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        int i = 0;
        for(; i < m; i++)
            dp[i][0] = 1;
        for(i = 0; i < n; i++)
            dp[0][i] = 1;
        for(i = 1; i < m; i++)
            for(int j = 1; j < n; j++)
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        return dp[m-1][n-1];
    }
}

二刷

二刷的时候还是用dp来做，这道题在DP中还是算作很简单的。

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