63. Unique Paths II - jiejackyzhang/leetcode-note GitHub Wiki

Follow up for "62. 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.

解题思路为Dynamic Programming。

与"62. Unique Paths"的区别在于将值为1的格子的path设为0，其余仍可按照p[i][j] = p[i-1][j] + p[i][j-1]进行计算。

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        dp[0][0] = 1 - obstacleGrid[0][0];
        for(int i = 1; i < m; i++) {
            dp[i][0] = obstacleGrid[i][0] == 1 ? 0 : dp[i-1][0];
        }
        for(int j = 1; j < n; j++) {
            dp[0][j] = obstacleGrid[0][j] == 1 ? 0 : dp[0][j-1];
        }
        for(int i = 1; i < m; i++) {
            for(int j = 1; j < n; j++) {
                dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
}

space 可以缩减为O(n)

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[] dp = new int[n];
        dp[0] = 1 - obstacleGrid[0][0];
        for(int j = 1; j < n; j++) {
            dp[j] = obstacleGrid[0][j] == 1 ? 0 : dp[j-1];
        }
        for(int i = 1; i < m; i++) {
            dp[0] = obstacleGrid[i][0] == 1 ? 0 : dp[0];
            for(int j = 1; j < n; j++) {
                dp[j] = obstacleGrid[i][j] == 1 ? 0 : dp[j] + dp[j-1];
            }
        }
        return dp[n-1];
    }
}
⚠️ **GitHub.com Fallback** ⚠️