【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?

【思路】

机器人每次向左或者向下走一步,在规则的长方形里有多少种方法到达终点。

动态规划,tmp[i][j]=tmp[i-1][j]+tmp[i][j-1],初始化状态只有一条路时只有以一种方法。

【代码】

class Solution {
  public int uniquePaths(int m, int n) {
    int [][]dp=new int[m+1][n+1];
    dp[0][0]=0;
    for(int i=0;i<m;i++){
      dp[i][0]=1;
    }
    for(int j=0;j<n;j++){
      dp[0][j]=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];
  }
}

 posted on 2018-11-07 19:46  alau  阅读(227)  评论(0编辑  收藏  举报