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飞过闲红千叶，夕岸在哪
类目归类
[LeetCode 63] Unique Paths II [Java]
02/08
1. Description
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.
Note: m and n will be at most 100.
2. 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.
3. Explanation
[tabby title="DP"]
Let dp[i][j] stands for the minimum sum of all numbers along its path when arrive grid (i,j)
(1). init first column
$$for(i : 0 \rightarrow M)$$
$$dp[i][0] = 1$$
(2). init first row
$$for(j : 0 \rightarrow N)$$
$$dp[0][j] = 1$$
(3). $$for(i : 1 \rightarrow M; j : 1 \rightarrow N)$$
$$dp[i][j] = dp[i-1][j] + dp[i][j-1] $$
[tabbyending]
4. Code
public class LeetCode0063 {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) {
return 0;
}
if (obstacleGrid[0][0] == 1) {
return 0;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
dp[0][0] = 1;
for (int i = 1; i < m; i++) {
if (obstacleGrid[i][0] == 1) {
dp[i][0] = 0;
} else {
dp[i][0] = dp[i - 1][0];
}
}
for (int j = 1; j < n; j++) {
if (obstacleGrid[0][j] == 1) {
dp[0][j] = 0;
} else {
dp[0][j] = dp[0][j - 1];
}
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 1) {
dp[i][j] = 0;
} else {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
public static void main(String[] args) {
LeetCode0063 leetcode = new LeetCode0063();
int[][] obstacleGrid = new int[][] { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };
System.out.println(leetcode.uniquePathsWithObstacles(obstacleGrid));
}
}
