62. Unique Paths
2016-06-04 22:42
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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?
![](http://leetcode.com/wp-content/uploads/2014/12/robot_maze.png)
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
DP解决,第一行所有格子只能来自它的左边,第一列所有格子只能来自它的上一个格子,这些格子的路径数均初始化为1。
递推方程 DP[i][j]=DP[i-1][j] + DP[i][j-1]
public static int uniquePaths(int m, int n)
{
int[][] dp=new int[m+1][n+1];
for(int i=1;i<=m;i++)
dp[i][1]=1;
for(int j=1;j<=n;j++)
dp[1][j]=1;
for(int i=2;i<=m;i++)
for(int j=2;j<=n;j++)
dp[i][j]=dp[i-1][j]+dp[i][j-1];
return dp[m]
;
}
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?
![](http://leetcode.com/wp-content/uploads/2014/12/robot_maze.png)
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
DP解决,第一行所有格子只能来自它的左边,第一列所有格子只能来自它的上一个格子,这些格子的路径数均初始化为1。
递推方程 DP[i][j]=DP[i-1][j] + DP[i][j-1]
public static int uniquePaths(int m, int n)
{
int[][] dp=new int[m+1][n+1];
for(int i=1;i<=m;i++)
dp[i][1]=1;
for(int j=1;j<=n;j++)
dp[1][j]=1;
for(int i=2;i<=m;i++)
for(int j=2;j<=n;j++)
dp[i][j]=dp[i-1][j]+dp[i][j-1];
return dp[m]
;
}
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