【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];
}
}