Unique Paths II

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本文介绍了一个算法问题的解决方案，该问题是求解在一个包含障碍物的网格中从起点到终点的不同路径数量。文章提供了详细的代码实现，采用动态规划的方法来计算可行路径的数量。

题目描述：

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.

代码如下：

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
		int m=obstacleGrid.length,n=obstacleGrid[0].length;
		int[][] matrix=new int[m][n];
		int sum=0;
		if(obstacleGrid[0][0]==1)
			return sum;
		matrix[0][0]=1;
		boolean flag=false;
		for(int i=1;i<n;i++){
			if(flag){
				matrix[0][i]=0;
			}else{
				if(obstacleGrid[0][i]==1){
					matrix[0][i]=0;
					flag=true;
				}else{
					matrix[0][i]=1;
				}
			}
		}
		flag=false;
		for(int i=1;i<m;i++){
			if(flag){
				matrix[i][0]=0;
			}else{
				if(obstacleGrid[i][0]==1){
					matrix[i][0]=0;
					flag=true;
				}else{
					matrix[i][0]=1;
				}
			}
		}
		for(int i=1;i<m;i++){
			for(int j=1;j<n;j++){
				if(obstacleGrid[i][j]==1){
					matrix[i][j]=0;
				}else{
					matrix[i][j]=matrix[i-1][j]+matrix[i][j-1];
				}
			}
		}
		return matrix[m-1][n-1];
    }
}