程序媛詹妮弗
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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 7 x 3 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

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

 

题目

给定MxN棋盘,只允许从左上往右下走,每次走一格。共有多少种走法?

 

思路

Matrix DP(二维DP) 问题

1. 初始化

预处理第一个row: dp[0][j] = 1  因为从左上起点出发,往右走的每一个unique path都是1

预处理第一个col:   dp[i][0]= 1  因为从左上起点出发,往下走的每一个unique path 都是1 

 

2. 转移方程

因为要求所有possible unique paths之和

dp[i][j] 要么来自dp[i-1][j] 要么来自dp[i][j-1]

 

代码

 1 class Solution {
 2     public int uniquePaths(int m, int n) {
 3         int[][]dp = new int[n][m]; // [row][col]
 4         
 5         // 预处理第一个col
 6         for(int i= 0; i < n; i++){
 7             dp[i][0] = 1;
 8         }  
 9         //预处理第一个row
10         for(int j=0; i < m; i++){
11             dp[0][j] = 1;
12         }
13         for(int i= 1; i < n; i++){
14             for(int j= 1; j < m; j++){
15                 dp[i][j] = dp[i-1][j] + dp[i][j-1];
16             }
17         }
18         return dp[n-1][m-1];
19     }
20 }

 

posted on 2019-03-05 16:29  程序媛詹妮弗  阅读(417)  评论(0编辑  收藏  举报