There is a robot on an m x n
grid. The robot is initially located at the top-left corner (i.e., grid[0][0]
). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]
). The robot can only move either down or right at any point in time.
Given the two integers m
and n
, return the number of possible unique paths that the robot can take to reach the bottom-right corner.
The test cases are generated so that the answer will be less than or equal to 2 * 109
.
Example 1:
Input: m = 3, n = 7 Output: 28
public int uniquePaths(int m, int n) {
Integer[][] map = new Integer[m][n];
for (int i = 0; i < m; i++)
map[i][0] = 1;
for (int j = 0; j < n; j++)
map[0][j]=1;
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++)
map[i][j] = map[i - 1][j] + map[i][j - 1]; //每一格路线的和都是其左格和上格的和
}
return map[m - 1][n - 1];
}
标签:Paths,right,map,int,robot,++,62,corner,Unique From: https://www.cnblogs.com/MarkLeeBYR/p/16886053.html