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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 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
一道简单的二维动态规划的问题,构造一个二维数组dp[][],当机器人走到(i,j)点的时候,因为机器人只能向右和向下两个方向走,这时总的路径就为dp[i][j] = dp[i - 1][j] + dp[i][j - 1],(i >= 1, j >= 1)。当i = 0 或j = 0的时候都只有一种走法,dp[0][j] = 1; dp[i][0] = 1,这是初始化。上面我们已经写出了递推式,这道题就解决了。代码如下:
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 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
一道简单的二维动态规划的问题,构造一个二维数组dp[][],当机器人走到(i,j)点的时候,因为机器人只能向右和向下两个方向走,这时总的路径就为dp[i][j] = dp[i - 1][j] + dp[i][j - 1],(i >= 1, j >= 1)。当i = 0 或j = 0的时候都只有一种走法,dp[0][j] = 1; dp[i][0] = 1,这是初始化。上面我们已经写出了递推式,这道题就解决了。代码如下:
public class Solution { public int uniquePaths(int m, int n) { int[][] dp = new int[m][n]; for(int i = 0; i < m; i++) dp[i][0] = 1; for(int i = 0; i < n; i++) dp[0][i] = 1; for(int i = 1; i < m; i++) for(int j = 1; j < n; j++) { dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; } return dp[m - 1][n - 1]; } }
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