[leetcode]63. Unique Paths II(有障碍物求路径和)

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).

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.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

 

题意

 

思路： 记忆化搜索(dfs + memo)

 

代码

public class _63_UniquePathsII {
    private int[][] memo;  // 缓存
    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;

        memo = new int[m][n];
        memo[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 memo[0][0]; // 回到起点，收敛条件

        if (memo[x][y] > 0) {
            return memo[x][y];
        } else {
            return memo[x][y] = dfs(obstacleGrid, x - 1, y) +
                    dfs(obstacleGrid, x, y - 1);
        }
    }
}