62. Unique Paths (JAVA)-优快云博客

博客讨论了在 m x n 网格中，位于左上角的机器人只能向下或向右移动，到达右下角的唯一路径数问题。还给出了 7 x 3 网格的示例，并提到可使用动态规划解决，状态与当前行和上一行有关，可用一维数组存储。

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

动态规划，当前值和左边格子与上边格子的状态有关；状态仅与当前行和上一行有关，并且由于是从上往下、从左往右遍历的，可以只使用一维数组存储。

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