一、题目

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.

二、题解

class Solution {
public:int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int m = obstacleGrid.size();int n = obstacleGrid[0].size();vector<vector<int>> dp(m,vector<int>(n,0));//初始化dp数组for(int i = 0;i < m;i++){if(obstacleGrid[i][0] == 0) dp[i][0] = 1;else break;}for(int j = 0;j < n;j++){if(obstacleGrid[0][j] == 0) dp[0][j] = 1;else break;}for(int i = 1;i < m;i++){for(int j = 1;j < n;j++){if(obstacleGrid[i][j] == 1) dp[i][j] = 0;else dp[i][j] = dp[i-1][j] + dp[i][j-1];}}return dp[m-1][n-1];}
};