63. 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 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.

这道题目的话也是使用相同的套路就好了，只是这道题目多了一个障碍物的条件，只需要在遇到障碍物的时候将到达障碍物格子的路径数设置为0就好了，没有什么难的

以下是程序：

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        int m = obstacleGrid.size();        if (m > 0) {            int n = obstacleGrid[0].size();            int** arr = new int*[m];            for (int i = 0; i < m; i++)                arr[i] = new int[n];            for (int i = 0; i < m; i++)                for (int j = 0; j < n; j++)                    arr[i][j] = 0;            for (int i = 0; i < n; i++) {                if (obstacleGrid[0][i] == 0) {                    arr[0][i] = 1;                } else {                    for (int j = i; j< n; j++)                        arr[0][j] = 0;                    break;                }            }            for (int i = 0; i < m; i++) {                if (obstacleGrid[i][0] == 0) {                    arr[i][0] = 1;                } else {                    for (int j = i; j < m; j++)                        arr[j][0] = 0;                    break;                }            }            for (int i = 1; i < m; i++) {                for (int j = 1; j < n; j++) {                    if (obstacleGrid[i][j] != 1)                        arr[i][j] = arr[i - 1][j] + arr[i][j - 1];                    else                        arr[i][j] = 0;                }            }            int num = arr[m - 1][n - 1];            for (int i = 0; i < m; i++)                delete []arr[i];            delete []arr;            return num;        } else {            return 0;        }    }};

这道题的解法的时间复杂度和空间复杂度都是O(m * n)