Unique Paths II

描述

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 3 × 3 grid as illustrated below.

  1. [
  2.   [0,0,0],
  3.   [0,1,0],
  4.   [0,0,0]
  5. ]

The total number of unique paths is 2.

Note: m and n will be at most 100.

备忘录法

在上一题的基础上改一下即可。相比动规，简单得多。

  1. // Unique Paths II
  2. // 深搜 + 缓存，即备忘录法
  3. public class Solution {
  4.     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
  5.         final int m = obstacleGrid.length;
  6.         final int n = obstacleGrid[0].length;
  7.         if (obstacleGrid[0][0] != 0 ||
  8.                 obstacleGrid[m - 1][n - 1] != 0) return 0;
  10.         f = new int[m][n];
  11.         f[0][0] = obstacleGrid[0][0] != 0 ? 0 : 1;
  12.         return dfs(obstacleGrid, m - 1, n - 1);
  13.     }
  15.     // @return 从 (0, 0) 到 (x, y) 的路径总数
  16.     int dfs(int[][] obstacleGrid, int x, int y) {
  17.         if (x < 0 || y < 0) return 0; // 数据非法，终止条件
  19.         // (x,y)是障碍
  20.         if (obstacleGrid[x][y] != 0) return 0;
  22.         if (x == 0 && y == 0) return f[0][0]; // 回到起点，收敛条件
  24.         if (f[x][y] > 0) {
  25.             return f[x][y];
  26.         } else {
  27.             return f[x][y] = dfs(obstacleGrid, x - 1, y) +
  28.                     dfs(obstacleGrid, x, y - 1);
  29.         }
  30.     }
  31.     private int[][] f;  // 缓存
  32. }

动规

与上一题类似，但要特别注意第一列的障碍。在上一题中，第一列全部是1，但是在这一题中不同，第一列如果某一行有障碍物，那么后面的行全为0。

  1. // Unique Paths II
  2. // 动规，滚动数组
  3. // 时间复杂度O(n^2)，空间复杂度O(n)
  4. public class Solution {
  5.     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
  6.         final int m = obstacleGrid.length;
  7.         final int n = obstacleGrid[0].length;
  8.         if (obstacleGrid[0][0] != 0 || 
  9.                 obstacleGrid[m-1][n-1] != 0) return 0;
  11.         int[] f = new int[n];
  12.         f[0] = obstacleGrid[0][0] != 0 ? 0 : 1;
  14.         for (int i = 0; i < m; i++) {
  15.             f[0] = f[0] == 0 ? 0 : (obstacleGrid[i][0] != 0 ? 0 : 1);
  16.             for (int j = 1; j < n; j++)
  17.                 f[j] = obstacleGrid[i][j] != 0 ? 0 : (f[j] + f[j - 1]);
  18.         }
  20.         return f[n - 1];
  21.     }
  22. }

相关题目

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/dfs/unique-paths-ii.html