Question:
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.
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int row = obstacleGrid.length;
int column = obstacleGrid[0].length;
int[][] totalPath = new int[row][column];
if (obstacleGrid[0][0] == 1) {
totalPath[0][0] = 0;
} else {
totalPath[0][0] = 1;
}
for (int i = 1; i < row; i++) {
if (obstacleGrid[i][0] == 1) {
totalPath[i][0] = 0;
} else {
totalPath[i][0] = totalPath[i - 1][0];
}
}
for (int j = 1; j < column; j++) {
if (obstacleGrid[0][j] == 1) {
totalPath[0][j] = 0;
} else {
totalPath[0][j] = totalPath[0][j - 1];
}
}
for (int i = 1; i < row; i++) {
for (int j = 1; j < column; j++) {
if (obstacleGrid[i][j] == 1) {
totalPath[i][j] = 0;
} else {
totalPath[i][j] = totalPath[i - 1][j] + totalPath[i][j - 1];
}
}
}
return totalPath[row-1][column-1];
}
}

本文探讨了在网格中加入障碍物后,独特的路径数量如何计算。通过实例展示了解决此类问题的方法。
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