【Leetcode】Unique Paths II-优快云博客

本文介绍了一种解决网格中存在障碍物时寻找从起点到终点的唯一路径数量的算法。通过动态规划的方法，考虑了障碍物的影响，并提供了两种实现方案，一种使用二维数组存储中间结果，另一种采用一维数组优化空间复杂度。

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.

Java:

这道题方法还是简单动态规划，只是增加了障碍判断，就是在遇到障碍的时候要设res=0表示走不通了，

直观的解法是用二维数组维护：如果第一行第一列有障碍，在初始化行的时候就跳出。

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

Reference:

1. http://blog.youkuaiyun.com/linhuanmars/article/details/22135231

空间复杂度一维O(n)

public int uniquePathsWithObstacles(int[][] obstacleGrid) {  
    if(obstacleGrid == null || obstacleGrid.length==0 || obstacleGrid[0].length==0)  
        return 0;  
    int[] res = new int[obstacleGrid[0].length];  
    res[0] = 1;  
    for(int i=0;i<obstacleGrid.length;i++)  
    {  
        for(int j=0;j<obstacleGrid[0].length;j++)  
        {  
            if(obstacleGrid[i][j]==1)  
            {  
                res[j]=0;  
            }  
            else  
            {  
                if(j>0)  
                    res[j] += res[j-1];  
            }  
        }  
    }  
    return res[obstacleGrid[0].length-1];  
}

2. http://www.cnblogs.com/feiling/p/3271548.html

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] steps = new int[m+2][n+2];
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                steps[i][j] = 0;
            }
        }
        steps[m][n+1] = 1;
        for(int i = m; i >= 1; i--){
            for(int j = n; j >= 1; j--){
                if(obstacleGrid[i - 1][j - 1] == 1){
                    steps[i][j] = 0;
                    continue;
                }
                steps[i][j] = steps[i+1][j] + steps[i][j+1];
            }
        }
        return steps[1][1];
    }