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0063. Unique Paths II
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‎markdown/0063. Unique Paths II.md

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### [63\. Unique Paths II](https://leetcode.com/problems/unique-paths-ii/)
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Difficulty: **Medium**
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A robot is located at the top-left corner of a _m_ x _n_ grid (marked 'Start' in the diagram below).
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The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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Now consider if some obstacles are added to the grids. How many unique paths would there be?
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![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)
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An obstacle and empty space is marked as `1` and `0` respectively in the grid.
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**Note:** _m_ and _n_ will be at most 100.
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**Example 1:**
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```
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Input:
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[
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  [0,0,0],
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  [0,1,0],
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  [0,0,0]
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]
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Output: 2
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Explanation:
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There is one obstacle in the middle of the 3x3 grid above.
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There are two ways to reach the bottom-right corner:
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1\. Right -> Right -> Down -> Down
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2\. Down -> Down -> Right -> Right
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```
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#### My Solution
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Language: **Java**
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```java
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class Solution {
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   public int uniquePathsWithObstacles(int[][] obstacleGrid) {
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       if (obstacleGrid == null) {
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           return 0;
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      }
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       int m = obstacleGrid.length;
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       if (m == 0) {
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           return 0;
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      }
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       int n = obstacleGrid[0].length;
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       int[][] ways = new int[m][n]; // 所有的格子都能到达，因此初始化为0，表示没有计算过。该数组用于保存到达该坐标的路径个数，防止重复计算
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       int i = uniquePathsWithObstacles(obstacleGrid, ways, m - 1, n - 1);
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       return i;
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  }
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​
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   private int uniquePathsWithObstacles(int[][] obstacleGrid, int[][] ways, int m, int n) {
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       if (m < 0 || n < 0 || obstacleGrid[m][n] == 1) {
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           return 0;
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      }
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       if (m + n == 0) {
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           ways[m][n] = 1;
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           return 1;
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      }
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       if (ways[m][n] > 0) {
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           return ways[m][n];
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      }
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       ways[m][n] = uniquePathsWithObstacles(obstacleGrid, ways, m, n - 1) + uniquePathsWithObstacles(obstacleGrid, ways, m - 1, n);
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       return ways[m][n];
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  }
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}
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```
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#### Recommend Solution
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Language: **Java**
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```java
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class Solution {
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public int uniquePathsWithObstacles(int[][] obstacleGrid) {
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int width = obstacleGrid[0].length;
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int[] dp = new int[width];
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dp[0] = 1;
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for (int[] row : obstacleGrid) {
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for (int j = 0; j < width; j++) {
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if (row[j] == 1)
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dp[j] = 0;
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else if (j > 0)
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dp[j] += dp[j - 1];
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}
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}
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return dp[width - 1];
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}
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// dp[j] += dp[j - 1];
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// is
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// dp[j] = dp[j] + dp[j - 1];
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// which is new dp[j] = old dp[j] + dp[j-1]
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// which is current cell = top cell + left cell
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}
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```
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![](https://image-hosting-1251780645.cos.ap-beijing.myqcloud.com/20190717000607.png)

‎src/main/java/leetcode/_63_/Main.java

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package leetcode._63_;
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/**
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* Created by zhangbo54 on 2019-03-04.
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*/
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public class Main {
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public static void main(String[] args) {
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Solution solution = new Solution();
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int[][] obstacleGrid = {{0, 1, 1}, {1, 1, 1}, {0, 0, 0}};
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System.out.println(solution.uniquePathsWithObstacles(obstacleGrid));
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int[][] obstacleGridw = {{0, 0, 0}, {0, 1, 0}, {0, 0, 0}};
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System.out.println(solution.uniquePathsWithObstacles(obstacleGridw));
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int[][] obstacleGridwq = {{0}};
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System.out.println(solution.uniquePathsWithObstacles(obstacleGridwq));
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int[][] obstacleGridwq2 = {{1}};
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System.out.println(solution.uniquePathsWithObstacles(obstacleGridwq2));
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}
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}
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‎src/main/java/leetcode/_63_/Solution.java

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package leetcode._63_;
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class Solution {
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public int uniquePathsWithObstacles(int[][] obstacleGrid) {
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if (obstacleGrid == null) {
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return 0;
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}
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int m = obstacleGrid.length;
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if (m == 0) {
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return 0;
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}
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int n = obstacleGrid[0].length;
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int[][] ways = new int[m][n]; // 所有的格子都能到达，因此初始化为0，表示没有计算过。该数组用于保存到达该坐标的路径个数，防止重复计算
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int i = uniquePathsWithObstacles(obstacleGrid, ways, m - 1, n - 1);
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return i;
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}
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private int uniquePathsWithObstacles(int[][] obstacleGrid, int[][] ways, int m, int n) {
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if (m < 0 || n < 0 || obstacleGrid[m][n] == 1) {
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return 0;
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}
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if (m + n == 0) {
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ways[m][n] = 1;
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return 1;
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}
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if (ways[m][n] > 0) {
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return ways[m][n];
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}
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ways[m][n] = uniquePathsWithObstacles(obstacleGrid, ways, m, n - 1) + uniquePathsWithObstacles(obstacleGrid, ways, m - 1, n);
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return ways[m][n];
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}
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}

‎src/main/java/leetcode/_63_/solution.md

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### [63\. Unique Paths II](https://leetcode.com/problems/unique-paths-ii/)
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Difficulty: **Medium**
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A robot is located at the top-left corner of a _m_ x _n_ grid (marked 'Start' in the diagram below).
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The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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Now consider if some obstacles are added to the grids. How many unique paths would there be?
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![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)
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An obstacle and empty space is marked as `1` and `0` respectively in the grid.
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**Note:** _m_ and _n_ will be at most 100.
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**Example 1:**
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```
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Input:
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[
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  [0,0,0],
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  [0,1,0],
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  [0,0,0]
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]
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Output: 2
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Explanation:
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There is one obstacle in the middle of the 3x3 grid above.
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There are two ways to reach the bottom-right corner:
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1\. Right -> Right -> Down -> Down
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2\. Down -> Down -> Right -> Right
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```
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#### My Solution
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Language: **Java**
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```java
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class Solution {
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   public int uniquePathsWithObstacles(int[][] obstacleGrid) {
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       if (obstacleGrid == null) {
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           return 0;
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      }
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       int m = obstacleGrid.length;
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       if (m == 0) {
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           return 0;
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      }
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       int n = obstacleGrid[0].length;
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       int[][] ways = new int[m][n]; // 所有的格子都能到达，因此初始化为0，表示没有计算过。该数组用于保存到达该坐标的路径个数，防止重复计算
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       int i = uniquePathsWithObstacles(obstacleGrid, ways, m - 1, n - 1);
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       return i;
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  }
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​
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   private int uniquePathsWithObstacles(int[][] obstacleGrid, int[][] ways, int m, int n) {
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       if (m < 0 || n < 0 || obstacleGrid[m][n] == 1) {
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           return 0;
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      }
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       if (m + n == 0) {
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           ways[m][n] = 1;
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           return 1;
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      }
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       if (ways[m][n] > 0) {
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           return ways[m][n];
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      }
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       ways[m][n] = uniquePathsWithObstacles(obstacleGrid, ways, m, n - 1) + uniquePathsWithObstacles(obstacleGrid, ways, m - 1, n);
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       return ways[m][n];
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  }
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}
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```
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#### Recommend Solution
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Language: **Java**
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```java
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class Solution {
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public int uniquePathsWithObstacles(int[][] obstacleGrid) {
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int width = obstacleGrid[0].length;
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int[] dp = new int[width];
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dp[0] = 1;
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for (int[] row : obstacleGrid) {
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for (int j = 0; j < width; j++) {
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if (row[j] == 1)
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dp[j] = 0;
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else if (j > 0)
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dp[j] += dp[j - 1];
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}
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}
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return dp[width - 1];
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}
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// dp[j] += dp[j - 1];
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// is
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// dp[j] = dp[j] + dp[j - 1];
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// which is new dp[j] = old dp[j] + dp[j-1]
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// which is current cell = top cell + left cell
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}
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```
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![](https://image-hosting-1251780645.cos.ap-beijing.myqcloud.com/20190717000607.png)

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