[LeetCode][Java] Unique Paths II

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題目:

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.

Note: m and n will be at most 100.

題意:

緊跟著題目《Unique Paths》,現給出這樣一題目:

假設在格子中加入一些障礙,會出現多少存在且唯一的不同路徑呢?

障礙和空白格子分別被標記為1 and 0 .

比方一個3x3的格子中的中間存在一個障礙,例如以下所看到的:

[  [0,0,0],  [0,1,0],  [0,0,0]]
總的路徑數為2.

演算法分析:

     思路與題目《Unique Paths》類似,不同之處為:

     初始化邊界上行和列時,出現障礙。後面路徑數dp的都是0

     中間的格子出現障礙時,該格子dp表示的路徑數直接填0

AC代碼:

public class Solution {    public int uniquePathsWithObstacles(int[][] obstacleGrid)     {        if(obstacleGrid==null||obstacleGrid.length==0)        return 0;    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)            dp[i][0] = 1;            else             break;        }        for(int j = 0; j < n; j++)        {        if(obstacleGrid[0][j]!=1)        dp[0][j] = 1;        else         break;        }        for(int i = 1; i < m; i++)        {            for(int j = 1; j< n; j++)            {            if(obstacleGrid[i][j]!=1)            dp[i][j] = dp[i-1][j] + dp[i][j-1];            else            dp[i][j]=0;            }        }        return dp[m-1][n-1];    }}


[LeetCode][Java] Unique Paths II

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