63. Unique Paths II

Problem:

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

思路：
添加额外的一行和一列，作为边界条件。然后起点变为(1,1)，且dp[1][1]=dp[0][1]+dp[1][0]=1。故设dp[0][1]或dp[1][0]为1即可。与62题相比增加一个判断条件，(i,j)是否有障碍，若无障碍则根据公式dp[i][j]=dp[i-1][j]+dp[i][j-1]求解。

Solution:

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

性能：
Runtime: 24 ms  Memory Usage: 9.9 MB