63. Unique Paths II

Posted 2020-11-10 妖域大都督

问题描述：

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

 

解题思路：

这道题跟Unique Path 的区别就是地图中有障碍物出现，此时这个格子无法被到达。

仍然使用动态规划进行求解，但是需要注意：

　　1. 当起点为障碍物时，无法到达终点

　　2. i = 0 和j = 0时不一定能够到达，取决于其上或左边能否被到达（错在了这里我：））

 

代码：

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