/** * 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]; } }
时间: 2024-10-25 22:09:27