Title: A robot is located in the upper left corner of an mxn grid (the starting point is marked as "Start" in the figure below).
The robot can only move one step down or to the right at a time. The robot tries to reach the bottom right corner of the grid (labeled "Finish" in the image below).
Now consider that there are obstacles in the grid. How many different paths will there be from the upper left corner to the lower right corner?
Obstacles and empty positions in the grid are represented by 1 and 0 respectively.
Note: The values of m and n do not exceed 100.
Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is an obstacle in the middle of the 3x3 grid.
There are 2 different paths from the upper left corner to the lower right corner:
- Right -> Right -> Down -> Down
- Down -> Down -> Right -> Right
Source: LeetCode
Link: https://leetcode-cn.com/problems/unique-paths-ii
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code show as below
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n= obstacleGrid[0].length;
if(obstacleGrid[0][0]==1||obstacleGrid[m-1][n-1]==1)
return 0;
int[][]dp = new int[m][n];
dp[0][0]=1;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(i==0&&j==0){
continue;
}else if(i==0){
if(obstacleGrid[i][j-1]!=1){
dp[i][j]=dp[i][j-1];
}else{
continue;
}
}else if(j==0){
if(obstacleGrid[i-1][j]!=1){
dp[i][j]=dp[i-1][j];
}else{
continue;
}
}else{
if(obstacleGrid[i-1][j]==1&&obstacleGrid[i][j-1]==1){
continue;
}else if(obstacleGrid[i-1][j]==1){
dp[i][j]=dp[i][j-1];
}else if(obstacleGrid[i][j-1]==1){
dp[i][j]=dp[i-1][j];
}else{
dp[i][j]=dp[i][j-1]+dp[i-1][j];
}
}
}
}
return dp[m-1][n-1];
}
}
According to dynamic programming, we find the transition equation as dp[i][j] = dp[i][j-1]+dp[i-1][j];
for the first row and first column dp[i][ j]=dp[i-1][j] or dp[i][j-1] In addition, we need to determine whether there is an obstacle in the corresponding position. If there is, skip the value of the position, and finally get the dp array The last element is the result