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.
一刷
題解：
也是DP問題倍宾，Unique Path一樣可以in place解決熙暴。要點(diǎn)是在設(shè)置第一行和第一列碰到obstacle的時候审丘，要將其以及之后的所有值設(shè)置為零制圈，因?yàn)闆]有路徑可以達(dá)到蒿往。之后在DP掃描矩陣的時候，也要講obstacle所在的位置清零芹枷。
Time Complexity - O(m * n)弟孟， Space Complexity - O(mn)汁咏。
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
for(int i=0; i<m; i++){
if(obstacleGrid[i][0] == 1){
for(int j=i; j<m; j++){
dp[j][0] = 0;
}
break;
}
else dp[i][0] = 1;
}
for(int i=0; i<n; i++){
if(obstacleGrid[0][i] == 1){
for(int j=i; j<n; j++){
dp[0][j] = 0;
}
break;
}
else dp[0][i] = 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{
if(i == 0||j==0) dp[i][j] = 1;
else dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
}
return dp[m-1][n-1];
}
}
