- 關(guān)鍵字：動(dòng)態(tài)規(guī)劃季二、遞歸
- 難度：Medium
- 題目大意：計(jì)算兩網(wǎng)格之間所有可能路徑五垮，只能向右或下行走
題目：
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、暴力遞歸：leetcode超時(shí)了；
2见秤、動(dòng)態(tài)規(guī)劃：定義p(i, j)為 從開始位置到坐標(biāo)(i，j)的所有可能路徑真椿，則：
- i==1或j==1時(shí)鹃答，顯然p(i, j) = 1;
- 其他情況下，p(i , j) = p(i-1, j) + p(i, j-1);
以m=7瀑粥，n=3為例挣跋，dp數(shù)組可以正序填充：
DP填充.jpeg
代碼實(shí)現(xiàn)：
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];
}
}
