63. Unique Paths II

难度: Medium

题目描述

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.

解题思路

代码实现

解决方案

java

 

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; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 && j == 0) {
                    if (obstacleGrid[0][0] == 1) {
                        return 0;
                    }
                    dp[i][j] = 1;
                    continue;
                }
                if (i == 0) {
                    // 直接根据当前元素是否为障碍物设置 dp 值
                    dp[i][j] = (obstacleGrid[i][j] == 1)? 0 : 1;
                    if(dp[i][j-1]==0){
                        dp[i][j]=0;
                    }
                    continue;
                }
                if (j == 0) {
                    // 直接根据当前元素是否为障碍物设置 dp 值
                    dp[i][j] = (obstacleGrid[i][j] == 1)? 0 : 1;
                    if(dp[i-1][j]==0){
                        dp[i][j]=0;
                    }
                    continue;
                }
                if (obstacleGrid[i][j] == 1) {
                    dp[i][j] = 0;
                    continue;
                }
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
}