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63. Unique Paths II

Difficulty: Medium

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

My Solution

Language: Java

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid == null) {
            return 0;
        }
        int m = obstacleGrid.length;
        if (m == 0) {
            return 0;
        }
        int n = obstacleGrid[0].length;
        int[][] ways = new int[m][n]; // 所有的格子都能到达，因此初始化为0，表示没有计算过。该数组用于保存到达该坐标的路径个数，防止重复计算
        int i = uniquePathsWithObstacles(obstacleGrid, ways, m - 1, n - 1);
        return i;
    }
​
    private int uniquePathsWithObstacles(int[][] obstacleGrid, int[][] ways, int m, int n) {
        if (m < 0 || n < 0 || obstacleGrid[m][n] == 1) {
            return 0;
        }
        if (m + n == 0) {
            ways[m][n] = 1;
            return 1;
        }
        if (ways[m][n] > 0) {
            return ways[m][n];
        }
        ways[m][n] = uniquePathsWithObstacles(obstacleGrid, ways, m, n - 1) + uniquePathsWithObstacles(obstacleGrid, ways, m - 1, n);
        return ways[m][n];
    }
}

Recommend Solution

Language: Java

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int width = obstacleGrid[0].length;
        int[] dp = new int[width];
        dp[0] = 1;
        for (int[] row : obstacleGrid) {
            for (int j = 0; j < width; j++) {
                if (row[j] == 1)
                    dp[j] = 0;
                else if (j > 0)
                    dp[j] += dp[j - 1];
            }
        }
        return dp[width - 1];
    }

    // dp[j] += dp[j - 1];
    // is
    // dp[j] = dp[j] + dp[j - 1];

    // which is new dp[j] = old dp[j] + dp[j-1]
    // which is current cell = top cell + left cell
}