Unique Paths II

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 3×3 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.

和Unique Paths I很像。

转移方程变一下即可，如果obstacleGrid[i][j] == 1 ，map[i][j] = 0;

否则 map[i][j] = map[i-1][j] + map[i][j-1];

这里需要注意数组的初始化。

int map[2][2] = {0};不能将所有元素初始化，需要一个循环来初始化。

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