LeetCode_OJ【63】Unique Paths II-优快云博客

本文链接：https://blog.youkuaiyun.com/hb_peng/article/details/50717985

本文介绍了一种使用动态规划解决网格中带有障碍物的唯一路径数量的方法。通过定义steps[][]数组来记录每个位置到达终点的可行解数量，并考虑障碍物的影响。

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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.

做了62题很显然这道题也要用动规来做，设置steps[][]数组来记录可行解的个数。steps[i][j]表示从(i,j)到(m-1,n-1)的可行解的个数。

显然steps[i][j] = steps[i+1][j] + steps[i][j+1]，当然这题要考虑障碍物的影响。

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] steps = new int[m][n];
        if(obstacleGrid[m-1][n-1] == 1)
        	return 0;
        else
        	steps[m-1][n-1] = 1;
        for(int i = m-1 ; i >= 0 ; i --){
        	for(int j = n-1 ; j >= 0 ; j--){
        		if(obstacleGrid[i][j] == 1)
        			continue;
        		if(i+1 < m && obstacleGrid[i+1][j] != 1)
        			steps[i][j] += steps[i+1][j];
        		if(j+1 < n && obstacleGrid[i][j+1] != 1)
        			steps[i][j] += steps[i][j+1];
        	}
        }
        return steps[0][0];
    }
}