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.
[
[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.
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。
// Unique Paths II
// 深搜 + 缓存，即备忘录法
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
final int m = obstacleGrid.length;
final int n = obstacleGrid[0].length;
if (obstacleGrid[0][0] != 0 ||
obstacleGrid[m - 1][n - 1] != 0) return 0;
f = new int[m][n];
f[0][0] = obstacleGrid[0][0] != 0 ? 0 : 1;
return dfs(obstacleGrid, m - 1, n - 1);
}
// @return 从 (0, 0) 到 (x, y) 的路径总数
int dfs(int[][] obstacleGrid, int x, int y) {
if (x < 0 || y < 0) return 0; // 数据非法，终止条件
// (x,y)是障碍
if (obstacleGrid[x][y] != 0) return 0;
if (x == 0 && y == 0) return f[0][0]; // 回到起点，收敛条件
if (f[x][y] > 0) {
return f[x][y];
} else {
return f[x][y] = dfs(obstacleGrid, x - 1, y) +
dfs(obstacleGrid, x, y - 1);
}
}
private int[][] f; // 缓存
}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。
// Unique Paths II
// 动规，滚动数组
// 时间复杂度O(n^2)，空间复杂度O(n)
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
final int m = obstacleGrid.length;
final int n = obstacleGrid[0].length;
if (obstacleGrid[0][0] != 0 ||
obstacleGrid[m-1][n-1] != 0) return 0;
int[] f = new int[n];
f[0] = obstacleGrid[0][0] != 0 ? 0 : 1;
for (int i = 0; i < m; i++) {
f[0] = f[0] == 0 ? 0 : (obstacleGrid[i][0] != 0 ? 0 : 1);
for (int j = 1; j < n; j++)
f[j] = obstacleGrid[i][j] != 0 ? 0 : (f[j] + f[j - 1]);
}
return f[n - 1];
}
}