63. Unique Paths II

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

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比起第I题 多了障碍物的判断,其实很简单就是多加个if而已. 不过本题有些边界条件导致不容易accept

首先是test case里面竟然有了0,0 位置和 m,n位置等于1 (有障碍物)的情况...我感觉这完全不make sense吧,本来就算出发点和结束点...

其次是有一个case 值特别大超过32位整数的范围,所以需要把vector的类型换成64位整数类型

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if(obstacleGrid.empty() || obstacleGrid[0].empty())return 0;
        int h=obstacleGrid.size();
        int w=obstacleGrid[0].size();
        vector<vector<uint64_t>> dp(h,vector<uint64_t>(w,0));
        dp[0][0]=1-obstacleGrid[0][0];
        for(int i=0;i<h;++i)
            for(int j=0;j<w;++j)
            {
                if(0==obstacleGrid[i][j])
                {
                    if(i) dp[i][j]+=dp[i-1][j];
                    if(j) dp[i][j]+=dp[i][j-1];
                }
            }
        return dp[h-1][w-1];
    }
};