Week 13 Unique Paths

最新推荐文章于 2018-12-10 15:45:09 发布

原创最新推荐文章于 2018-12-10 15:45:09 发布 · 133 阅读

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本文探讨了一个经典的动态规划问题——寻找网格中从左上角到右下角的不同路径数量。通过分析机器人仅能向下或向右移动的条件，我们建立了状态方程，并提供了详细的算法实现，包括初始化边界条件和迭代计算过程。

62.Unique Paths

问题概述

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?
在这里插入图片描述

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

分析

这是一道基本动态规划问题。
机器人只能向下或向右移动，当它位于一个坐标点时，有两种可能：

1.It arrives at that point from above (moving down to that point);
2.It arrives at that point from left (moving right to that point).

假设到达一个点（i，j）的路径数用P[i][j]表示，
因而可以建立状态方程：
P[i][j]=P[i-1][j]+P[i][j-1]
P[0][j] = 1, P[i][0] = 1（初态）

代码

/c++/
class Solution {
    int uniquePaths(int m, int n) {
        vector<vector<int> > path(m, vector<int> (n, 1));
        for (int i = 1; i < m; i++)
            for (int j = 1; j < n; j++)
                path[i][j] = path[i - 1][j] + path[i][j - 1];
        return path[m - 1][n - 1];
    }
};