[LeetCode] 63. Unique Paths II

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

Example 1:

Input: obstacleGrid = [[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

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

Constraints:

• m == obstacleGrid.length
• n == obstacleGrid[i].length
• 1 <= m, n <= 100
• obstacleGrid[i][j] is 0 or 1.

不同路径 II。

一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为 “Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为 “Finish”）。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？

网格中的障碍物和空位置分别用 1 和 0 来表示。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/unique-paths-ii
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题意跟版本一一样，唯一的不同点在于路径上会有障碍物。依然是返回到底有多少不同的路径能从左上角走到右下角。我还是提供自上而下和自下而上两种解法。

DP自上而下

时间O(mn)

空间O(mn)

Java实现

class Solution {
    int[][] memo;
    
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        memo = new int[m][n];
        return dp(obstacleGrid, m - 1, n - 1);
    }
    
    private int dp(int[][] grid, int i, int j) {
        // base case
        // 越界或者遇到障碍物，直接返回0
        if (i < 0 || j < 0 || grid[i][j] == 1) {
            return 0;
        }
        if (i == 0 && j == 0) {
            return 1;
        }
        if (memo[i][j] > 0) {
            return memo[i][j];
        }
        memo[i][j] = dp(grid, i - 1, j) + dp(grid, i, j - 1);
        return memo[i][j];
    }
}

// dp自上而下

 

DP自下而上

时间O(mn)

空间O(mn)

Java实现

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

// dp自下而上

 

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