/**
* Created by lvhao on 2017/7/6.
* 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?
动态规划,目标是到最终点的路径有几条。每个点的路径都是能到这个点的上一个点的路径之和,由于
只能向下和向右走,所以动态方程是A[i][j] = A[i][j-1] + A[j][j+1]
算出每个点的值就行,最后返回A[m-1][n-1]
*/
public class Q62UniquePaths {
public static void main(String[] args) {
System.out.println(uniquePaths(3,3));
}
public static int uniquePaths(int m, int n) {
if (m == 1 || n == 1)
return 1;
int[][] res = new int[m][n];
//一开始要先把二维数组上和左边的初始化为1,这是初始条件
for (int i = 0; i < m; i++) {
res[i][0] = 1;
}
for (int i = 0; i < n; i++) {
res[0][i] = 1;
}
//动态规划
for (int i = 1; i < m; i++) {
for (int j = 1;j < n;j++)
{
res[i][j] = res[i-1][j] + res[i][j-1];
}
}
return res[m-1][n-1];
}
}