LeetCode063. Unique Paths II

Posted 2020-09-17 Vincent丶

题目：

Follow up for "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.

题解：

Solution 1 ()(未优化）

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

Solution 2 ()

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