Week 13 Unique Paths
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?
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];
}
};