题目如下:
Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middle of a 3x3 grid as illustrated below. [ [0,0,0], [0,1,0], [0,0,0] ] The total number of unique paths is 2. Note: m and n will be at most 100.
思路:
同 LeetCode 62. Unique Paths, 只不过决定每个res[i][j]的时候,都要先判断:obstacleGrid[i][j]如果等于1,则res[i][j]=0; 否则正常进行。
本题代码:
import java.util.Scanner; /** * Created by yuanxu on 17/4/13. */ public class DP63 { public static int uniquePathsWithObstacles(int[][] obstacleGrid) { int m = obstacleGrid.length; int n = obstacleGrid[0].length; int res[][] = new int[m][n];// 数据从0开始 // boundary condition res[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1; for (int i=1; i<m; i++) { if (res[i-1][0] == 0) { res[i][0] = 0; } else if (obstacleGrid[i][0] == 1) { res[i][0] = 0; } else { res[i][0] = 1; } } for (int j=1; j<n; j++) { if (res[0][j-1] == 0) { res[0][j-1] = 0; } else if (obstacleGrid[0][j] == 1) { res[0][j] = 0; } else { res[0][j] = 1; } } // DP for (int i=1; i<m; i++) { for (int j=1; j<n; j++) { if (obstacleGrid[i][j] == 1) { res[i][j] = 0; } else { res[i][j] = res[i-1][j] + res[i][j-1]; } } } return res[m-1][n-1]; } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); // int obstacleGrid[][] = {{0,0,0},{0,1,0},{0,0,0}}; // int obstacleGrid[][] = {{1}}; int obstacleGrid[][] = {{0,0},{0,1}}; System.out.println(uniquePathsWithObstacles(obstacleGrid)); } }