Unique Paths II
package cn.edu.xidian.sselab.array;
/**
*
* @author zhiyong wang
* title: Unique Paths II
* content:
* 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.
*
*/
public class UniquePathsII {
//递归没有想出怎么做来,不过这个是类似PascalTriangle
/**
* 1 1 1 1
* 1 2 3 4
* 1 3 6 10
* 1 4 10 20
* 即4*4的表,最多有20条路线,如果有障碍的话,把障碍换成0,把非障碍换成1
* 自己没有想出解决办法,最后看的答案:思路是这样的,首先判断(0,0)的值,然后初始化第一行与第一列的值,
* 第一个值遇0变1,遇1变0,对行进行初始化的时候,如果当前值是1,则置为0,如果是0,则前一个值赋予它
* 对列 进行初始化的时候也是,当前值是1,则置为0,如果是0,则将上一个值赋予它
* 初始化完成之后,就是对剩余所有的值开始进行判断,如果遇到1,则置为0,如果遇到0,则去上面值与前面值的和作为该点的值
*/
public int uniquePathsWithObstacles(int[][] obstacleGrid){
int length = obstacleGrid.length;
int width = obstacleGrid[0].length;
//第一步先初始化第一行与第一列
obstacleGrid[0][0] ^= 1;
for(int i=1;i<width;i++)
obstacleGrid[0][i] = obstacleGrid[0][i]==1?0:obstacleGrid[0][i-1];
for(int i=1;i<length;i++)
obstacleGrid[i][0] = obstacleGrid[i][0]==1?0:obstacleGrid[i-1][0];
//第二步开始求最后路径的和
for(int i=1;i<length;i++){
for(int j=1;j<width;j++){
obstacleGrid[i][j] = obstacleGrid[i][j]==1?0:obstacleGrid[i-1][j]+obstacleGrid[i][j-1];
}
}
return obstacleGrid[length-1][width-1];
}
}