LeetCode进阶之路（Unique Paths）_robot navigation a robot needs move from top left -优快云博客

本文链接：https://blog.youkuaiyun.com/yulingkai88/article/details/52530889

本文介绍了一个经典的动态规划问题——计算从起点到终点的所有唯一路径数量。机器人只能向下或向右移动，通过动态规划的方法求解路径总数。

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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 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

题意：从start位置开始，只能向下或者向右，有多少条路径可以达到Finish。

思路：用动态规划的思路，例如最后要达到Finish的位置，有两个位置可以达到，那把这两个位置的数目加起来就是总的条数。这样往上推倒即可。

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

PS：好一段时间没更新了，在北京出差实在是忙的没时间，这个星期回到杭州，继续保持更新，坚持就是胜利。