题目描述
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.
分析
有障碍物的时候(obstacleGrid[i][j]==1),将对应的dp置0(dp[i][j]=0)。
代码
public static int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null || obstacleGrid[0] == null) {
return 0;
}
if (obstacleGrid[0][0]==1) {
return 0;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
for (int y = 1; y < n; y++) {
if (obstacleGrid[0][y] == 0) {
dp[0][y] = 1;
} else {
break;
}
}
for (int x = 0; x < m; x++) {
if (obstacleGrid[x][0] == 0) {
dp[x][0] = 1;
} else {
break;
}
}
for (int y = 1; y < n; y++) {
for (int x = 1; x < m; x++) {
if (obstacleGrid[x][y] == 1) {
dp[x][y] = 0;
} else {
dp[x][y] = dp[x - 1][y] + dp[x][y - 1];
}
}
}
return dp[m-1][n-1];
}