Unique Paths II

Posted 2020-10-01 apanda009

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.

 Notice

m and n will be at most 100.

Have you met this question in a real interview? Yes
Example
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.

Tags 

初始化, if 判断不同的状态

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // write your code here
        if (obstacleGrid == null || obstacleGrid.length == 0) {
            return -1;
        }
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        //state
        int[][] sum = new int[m][n];
        //initialize
        if (obstacleGrid[0][0] == 1) {
            sum[0][0] = 0;
        } else {
            sum[0][0] = 1;
        }
        for (int i = 1; i < n; i++) {
            if (obstacleGrid[0][i] == 1 || sum[0][i - 1] == 0) {
                sum[0][i] = 0;
            } else {
                sum[0][i] = 1;
            }
        }
        for (int i = 1; i < m; i++) {
            if (obstacleGrid[i][0] == 1 || sum[i - 1][0] == 0) {
                sum[i][0] = 0;
            } else {
                sum[i][0] = 1;
            }
        }
        
        //function
        for (int i = 1; i < m; i++) {
            
           for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    sum[i][j] = 0;
                } else {
                    sum[i][j] = sum[i - 1][j] + sum[i][j - 1];
                }
            }
        }
        //ans
        return sum[m - 1][n - 1];
        }