- 关键字：动态规划、递归
- 难度：Medium
- 题目大意：计算两网格之间所有可能路径，只能向右或下行走
题目：
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?
robot_maze.png
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:
- Right -> Right -> Down
- Right -> Down -> Right
- Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
思路：
1、暴力递归：leetcode超时了；
2、动态规划：定义p(i, j)为 从开始位置到坐标(i，j)的所有可能路径，则：
- i==1或j==1时，显然p(i, j) = 1;
- 其他情况下，p(i , j) = p(i-1, j) + p(i, j-1);
以m=7，n=3为例，dp数组可以正序填充：
DP填充.jpeg
代码实现：
解法1：
public class solution {
public int uniquePaths(int m, int n) {
return process(n,m);
}
private int process(int i, int j) {
if(i==1||j==1) return 1;
return process(i-1,j) + process(i,j-1);
}
}
解法2：
class Solution {
public int uniquePaths(int m, int n) {
int[][] dp = new int[n+1][m+1];
for(int i=1;i<n+1;i++) {
for(int j=1;j<m+1;j++) {
if(i==1||j==1){
dp[i][j] = 1;
}
}
}
for(int i=2;i<n+1;i++) {
for(int j=2;j<m+1;j++) {
dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
return dp[n][m];
}
}
