摘要:和完全一样的做法,只要在初始化首行和首列遇到时置零且即可。对了,数组其它元素遇到也要置零喏,不过就不要啦。
Problem
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.
Noticem and n will be at most 100.
ExampleFor 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和Unique Path完全一样的做法,只要在初始化首行和首列遇到obstacle时置零且break即可。对了,数组其它元素遇到obstacle也要置零喏,不过就不要break啦。
Solution二维DP
public class Solution { public int uniquePathsWithObstacles(int[][] A) { int m = A.length, n = A[0].length; int[][] dp = new int[m][n]; for (int i = 0; i < m; i++) { if (A[i][0] == 0) dp[i][0] = 1; else break; } for (int j = 0; j < n; j++) { if (A[0][j] == 0) dp[0][j] = 1; else break; } for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { if (A[i][j] == 0) dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m-1][n-1]; } }
一维DP
public class Solution { public int uniquePathsWithObstacles(int[][] A) { int n = A[0].length; int[] dp = new int[n]; dp[0] = 1; for (int i = 0; i < A.length; i++) { for (int j = 0; j < n; j++) { if (A[i][j] == 1) dp[j] = 0; else if (j > 0) dp[j] += dp[j-1]; } } return dp[n-1]; } }
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