742 lines
21 KiB
Markdown
742 lines
21 KiB
Markdown
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<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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# 63. 不同路径 II
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[力扣题目链接](https://leetcode.cn/problems/unique-paths-ii/)
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一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。
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机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。
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现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？
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网格中的障碍物和空位置分别用 1 和 0 来表示。
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示例 1：
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* 输入：obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
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* 输出：2
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解释：
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* 3x3 网格的正中间有一个障碍物。
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* 从左上角到右下角一共有 2 条不同的路径：
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1. 向右 -> 向右 -> 向下 -> 向下
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2. 向下 -> 向下 -> 向右 -> 向右
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示例 2：
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* 输入：obstacleGrid = [[0,1],[0,0]]
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* 输出：1
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提示：
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* m == obstacleGrid.length
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* n == obstacleGrid[i].length
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* 1 <= m, n <= 100
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* obstacleGrid[i][j] 为 0 或 1
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# 算法公开课
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**《代码随想录》算法视频公开课：[动态规划，这次遇到障碍了| LeetCode：63. 不同路径 II](https://www.bilibili.com/video/BV1Ld4y1k7c6/)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。
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## 思路
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这道题相对于[62.不同路径](https://programmercarl.com/0062.不同路径.html) 就是有了障碍。
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第一次接触这种题目的同学可能会有点懵，这有障碍了，应该怎么算呢？
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[62.不同路径](https://programmercarl.com/0062.不同路径.html)中我们已经详细分析了没有障碍的情况，有障碍的话，其实就是标记对应的dp table（dp数组）保持初始值(0)就可以了。
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动规五部曲：
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1. 确定dp数组（dp table）以及下标的含义
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dp[i][j] ：表示从（0 ，0）出发，到(i, j) 有dp[i][j]条不同的路径。
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2. 确定递推公式
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递推公式和62.不同路径一样，dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。
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但这里需要注意一点，因为有了障碍，(i, j)如果就是障碍的话应该就保持初始状态（初始状态为0）。
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所以代码为：
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```cpp
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if (obstacleGrid[i][j] == 0) { // 当(i, j)没有障碍的时候，再推导dp[i][j]
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dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
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}
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```
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3. dp数组如何初始化
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在[62.不同路径](https://programmercarl.com/0062.不同路径.html)不同路径中我们给出如下的初始化：
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```cpp
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vector<vector<int>> dp(m, vector<int>(n, 0)); // 初始值为0
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for (int i = 0; i < m; i++) dp[i][0] = 1;
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for (int j = 0; j < n; j++) dp[0][j] = 1;
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```
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因为从(0, 0)的位置到(i, 0)的路径只有一条，所以dp[i][0]一定为1，dp[0][j]也同理。
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但如果(i, 0) 这条边有了障碍之后，障碍之后（包括障碍）都是走不到的位置了，所以障碍之后的dp[i][0]应该还是初始值0。
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如图：
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下标(0, j)的初始化情况同理。
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所以本题初始化代码为：
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```CPP
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vector<vector<int>> dp(m, vector<int>(n, 0));
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for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;
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for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0][j] = 1;
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```
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**注意代码里for循环的终止条件，一旦遇到obstacleGrid[i][0] == 1的情况就停止dp[i][0]的赋值1的操作，dp[0][j]同理**
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4. 确定遍历顺序
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从递归公式dp[i][j] = dp[i - 1][j] + dp[i][j - 1] 中可以看出，一定是从左到右一层一层遍历，这样保证推导dp[i][j]的时候，dp[i - 1][j] 和 dp[i][j - 1]一定是有数值。
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代码如下：
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```CPP
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for (int i = 1; i < m; i++) {
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for (int j = 1; j < n; j++) {
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if (obstacleGrid[i][j] == 1) continue;
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dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
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}
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}
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```
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5. 举例推导dp数组
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拿示例1来举例如题：
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对应的dp table 如图：
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如果这个图看不懂，建议再理解一下递归公式，然后照着文章中说的遍历顺序，自己推导一下！
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动规五部分分析完毕，对应C++代码如下：
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```CPP
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class Solution {
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public:
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int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
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int m = obstacleGrid.size();
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int n = obstacleGrid[0].size();
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if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) //如果在起点或终点出现了障碍，直接返回0
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return 0;
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vector<vector<int>> dp(m, vector<int>(n, 0));
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for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;
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for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0][j] = 1;
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for (int i = 1; i < m; i++) {
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for (int j = 1; j < n; j++) {
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if (obstacleGrid[i][j] == 1) continue;
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dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
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}
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}
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return dp[m - 1][n - 1];
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}
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};
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```
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* 时间复杂度：O(n × m)，n、m 分别为obstacleGrid 长度和宽度
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* 空间复杂度：O(n × m)
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同样我们给出空间优化版本：
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```CPP
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class Solution {
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public:
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int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
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if (obstacleGrid[0][0] == 1)
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return 0;
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vector<int> dp(obstacleGrid[0].size());
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for (int j = 0; j < dp.size(); ++j)
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if (obstacleGrid[0][j] == 1)
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dp[j] = 0;
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else if (j == 0)
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dp[j] = 1;
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else
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dp[j] = dp[j-1];
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for (int i = 1; i < obstacleGrid.size(); ++i)
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for (int j = 0; j < dp.size(); ++j){
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if (obstacleGrid[i][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] + dp[j-1];
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}
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return dp.back();
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}
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};
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```
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* 时间复杂度：O(n × m)，n、m 分别为obstacleGrid 长度和宽度
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* 空间复杂度：O(m)
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## 总结
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本题是[62.不同路径](https://programmercarl.com/0062.不同路径.html)的障碍版，整体思路大体一致。
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但就算是做过62.不同路径，在做本题也会有感觉遇到障碍无从下手。
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其实只要考虑到，遇到障碍dp[i][j]保持0就可以了。
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也有一些小细节，例如：初始化的部分，很容易忽略了障碍之后应该都是0的情况。
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## 其他语言版本
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### 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 m = obstacleGrid.length;
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int n = obstacleGrid[0].length;
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int[][] dp = new int[m][n];
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//如果在起点或终点出现了障碍，直接返回0
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if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {
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return 0;
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}
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for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) {
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dp[i][0] = 1;
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}
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for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
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dp[0][j] = 1;
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}
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for (int i = 1; i < m; i++) {
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for (int j = 1; j < n; j++) {
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dp[i][j] = (obstacleGrid[i][j] == 0) ? dp[i - 1][j] + dp[i][j - 1] : 0;
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}
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}
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return dp[m - 1][n - 1];
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}
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}
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```
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```java
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// 空间优化版本
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class Solution {
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public int uniquePathsWithObstacles(int[][] obstacleGrid) {
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int m = obstacleGrid.length;
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int n = obstacleGrid[0].length;
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int[] dp = new int[n];
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for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
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dp[j] = 1;
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}
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for (int i = 1; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (obstacleGrid[i][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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}
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return dp[n - 1];
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}
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}
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```
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### Python
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动态规划（版本一）
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```python
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class Solution:
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def uniquePathsWithObstacles(self, obstacleGrid):
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m = len(obstacleGrid)
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n = len(obstacleGrid[0])
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if obstacleGrid[m - 1][n - 1] == 1 or obstacleGrid[0][0] == 1:
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return 0
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dp = [[0] * n for _ in range(m)]
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for i in range(m):
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if obstacleGrid[i][0] == 0: # 遇到障碍物时，直接退出循环，后面默认都是0
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dp[i][0] = 1
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else:
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break
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for j in range(n):
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if obstacleGrid[0][j] == 0:
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dp[0][j] = 1
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else:
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break
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for i in range(1, m):
|
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for j in range(1, n):
|
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if obstacleGrid[i][j] == 1:
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continue
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dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
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return dp[m - 1][n - 1]
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```
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动态规划（版本二）
|
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```python
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class Solution:
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def uniquePathsWithObstacles(self, obstacleGrid):
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m = len(obstacleGrid) # 网格的行数
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n = len(obstacleGrid[0]) # 网格的列数
|
||
|
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if obstacleGrid[m - 1][n - 1] == 1 or obstacleGrid[0][0] == 1:
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# 如果起点或终点有障碍物，直接返回0
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return 0
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|
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dp = [[0] * n for _ in range(m)] # 创建一个二维列表用于存储路径数
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# 设置起点的路径数为1
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dp[0][0] = 1 if obstacleGrid[0][0] == 0 else 0
|
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|
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# 计算第一列的路径数
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for i in range(1, m):
|
||
if obstacleGrid[i][0] == 0:
|
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dp[i][0] = dp[i - 1][0]
|
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|
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# 计算第一行的路径数
|
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for j in range(1, n):
|
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if obstacleGrid[0][j] == 0:
|
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dp[0][j] = dp[0][j - 1]
|
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|
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# 计算其他位置的路径数
|
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for i in range(1, m):
|
||
for j in range(1, n):
|
||
if obstacleGrid[i][j] == 1:
|
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continue
|
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dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
|
||
|
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return dp[m - 1][n - 1] # 返回终点的路径数
|
||
|
||
|
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```
|
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动态规划（版本三）
|
||
|
||
```python
|
||
class Solution:
|
||
def uniquePathsWithObstacles(self, obstacleGrid):
|
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if obstacleGrid[0][0] == 1:
|
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return 0
|
||
|
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dp = [0] * len(obstacleGrid[0]) # 创建一个一维列表用于存储路径数
|
||
|
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# 初始化第一行的路径数
|
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for j in range(len(dp)):
|
||
if obstacleGrid[0][j] == 1:
|
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dp[j] = 0
|
||
elif j == 0:
|
||
dp[j] = 1
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||
else:
|
||
dp[j] = dp[j - 1]
|
||
|
||
# 计算其他行的路径数
|
||
for i in range(1, len(obstacleGrid)):
|
||
for j in range(len(dp)):
|
||
if obstacleGrid[i][j] == 1:
|
||
dp[j] = 0
|
||
elif j != 0:
|
||
dp[j] = dp[j] + dp[j - 1]
|
||
|
||
return dp[-1] # 返回最后一个元素，即终点的路径数
|
||
|
||
```
|
||
动态规划（版本四）
|
||
|
||
```python
|
||
class Solution:
|
||
def uniquePathsWithObstacles(self, obstacleGrid):
|
||
if obstacleGrid[0][0] == 1:
|
||
return 0
|
||
|
||
m, n = len(obstacleGrid), len(obstacleGrid[0])
|
||
|
||
dp = [0] * n # 创建一个一维列表用于存储路径数
|
||
|
||
# 初始化第一行的路径数
|
||
for j in range(n):
|
||
if obstacleGrid[0][j] == 1:
|
||
break
|
||
dp[j] = 1
|
||
|
||
# 计算其他行的路径数
|
||
for i in range(1, m):
|
||
if obstacleGrid[i][0] == 1:
|
||
dp[0] = 0
|
||
for j in range(1, n):
|
||
if obstacleGrid[i][j] == 1:
|
||
dp[j] = 0
|
||
else:
|
||
dp[j] += dp[j - 1]
|
||
|
||
return dp[-1] # 返回最后一个元素，即终点的路径数
|
||
|
||
```
|
||
|
||
动态规划（版本五）
|
||
|
||
```python
|
||
class Solution:
|
||
def uniquePathsWithObstacles(self, obstacleGrid):
|
||
if obstacleGrid[0][0] == 1:
|
||
return 0
|
||
|
||
m, n = len(obstacleGrid), len(obstacleGrid[0])
|
||
|
||
dp = [0] * n # 创建一个一维列表用于存储路径数
|
||
|
||
# 初始化第一行的路径数
|
||
for j in range(n):
|
||
if obstacleGrid[0][j] == 1:
|
||
break
|
||
dp[j] = 1
|
||
|
||
# 计算其他行的路径数
|
||
for i in range(1, m):
|
||
if obstacleGrid[i][0] == 1:
|
||
dp[0] = 0
|
||
for j in range(1, n):
|
||
if obstacleGrid[i][j] == 1:
|
||
dp[j] = 0
|
||
continue
|
||
|
||
dp[j] += dp[j - 1]
|
||
|
||
return dp[-1] # 返回最后一个元素，即终点的路径数
|
||
|
||
|
||
```
|
||
### Go
|
||
|
||
```go
|
||
func uniquePathsWithObstacles(obstacleGrid [][]int) int {
|
||
m, n := len(obstacleGrid), len(obstacleGrid[0])
|
||
//如果在起点或终点出现了障碍，直接返回0
|
||
if obstacleGrid[m-1][n-1] == 1 || obstacleGrid[0][0] == 1 {
|
||
return 0
|
||
}
|
||
// 定义一个dp数组
|
||
dp := make([][]int, m)
|
||
for i, _ := range dp {
|
||
dp[i] = make([]int, n)
|
||
}
|
||
// 初始化, 如果是障碍物, 后面的就都是0, 不用循环了
|
||
for i := 0; i < m && obstacleGrid[i][0] == 0; i++ {
|
||
dp[i][0] = 1
|
||
}
|
||
for i := 0; i < n && obstacleGrid[0][i] == 0; i++ {
|
||
dp[0][i] = 1
|
||
}
|
||
// dp数组推导过程
|
||
for i := 1; i < m; i++ {
|
||
for j := 1; j < n; j++ {
|
||
// 如果obstacleGrid[i][j]这个点是障碍物, 那么dp[i][j]保持为0
|
||
if obstacleGrid[i][j] != 1 {
|
||
// 否则我们需要计算当前点可以到达的路径数
|
||
dp[i][j] = dp[i-1][j] + dp[i][j-1]
|
||
}
|
||
}
|
||
}
|
||
return dp[m-1][n-1]
|
||
}
|
||
```
|
||
|
||
### Javascript
|
||
|
||
```Javascript
|
||
var uniquePathsWithObstacles = function(obstacleGrid) {
|
||
const m = obstacleGrid.length
|
||
const n = obstacleGrid[0].length
|
||
const dp = Array(m).fill().map(item => Array(n).fill(0))
|
||
|
||
for (let i = 0; i < m && obstacleGrid[i][0] === 0; ++i) {
|
||
dp[i][0] = 1
|
||
}
|
||
|
||
for (let i = 0; i < n && obstacleGrid[0][i] === 0; ++i) {
|
||
dp[0][i] = 1
|
||
}
|
||
|
||
for (let i = 1; i < m; ++i) {
|
||
for (let j = 1; j < n; ++j) {
|
||
dp[i][j] = obstacleGrid[i][j] === 1 ? 0 : dp[i - 1][j] + dp[i][j - 1]
|
||
}
|
||
}
|
||
|
||
return dp[m - 1][n - 1]
|
||
};
|
||
|
||
// 版本二：内存优化，直接以原数组为dp数组
|
||
var uniquePathsWithObstacles = function(obstacleGrid) {
|
||
const m = obstacleGrid.length;
|
||
const n = obstacleGrid[0].length;
|
||
for (let i = 0; i < m; i++) {
|
||
for (let j = 0; j < n; j++) {
|
||
if (obstacleGrid[i][j] === 0) {
|
||
// 不是障碍物
|
||
if (i === 0) {
|
||
// 取左边的值
|
||
obstacleGrid[i][j] = obstacleGrid[i][j - 1] ?? 1;
|
||
} else if (j === 0) {
|
||
// 取上边的值
|
||
obstacleGrid[i][j] = obstacleGrid[i - 1]?.[j] ?? 1;
|
||
} else {
|
||
// 取左边和上边的和
|
||
obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
|
||
}
|
||
} else {
|
||
// 如果是障碍物，则路径为0
|
||
obstacleGrid[i][j] = 0;
|
||
}
|
||
}
|
||
}
|
||
return obstacleGrid[m - 1][n - 1];
|
||
};
|
||
```
|
||
|
||
|
||
|
||
### TypeScript
|
||
|
||
```typescript
|
||
function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
|
||
/**
|
||
dp[i][j]: 到达(i, j)的路径数
|
||
dp[0][*]: 用u表示第一个障碍物下标，则u之前为1，u之后（含u）为0
|
||
dp[*][0]: 同上
|
||
...
|
||
dp[i][j]: obstacleGrid[i][j] === 1 ? 0 : dp[i-1][j] + dp[i][j-1];
|
||
*/
|
||
const m: number = obstacleGrid.length;
|
||
const n: number = obstacleGrid[0].length;
|
||
const dp: number[][] = new Array(m).fill(0).map(_ => new Array(n).fill(0));
|
||
for (let i = 0; i < m && obstacleGrid[i][0] === 0; i++) {
|
||
dp[i][0] = 1;
|
||
}
|
||
for (let i = 0; i < n && obstacleGrid[0][i] === 0; i++) {
|
||
dp[0][i] = 1;
|
||
}
|
||
for (let i = 1; i < m; i++) {
|
||
for (let j = 1; j < n; j++) {
|
||
if (obstacleGrid[i][j] === 1) continue;
|
||
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
|
||
}
|
||
}
|
||
return dp[m - 1][n - 1];
|
||
};
|
||
```
|
||
|
||
### Rust
|
||
|
||
```Rust
|
||
impl Solution {
|
||
pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
|
||
let m: usize = obstacle_grid.len();
|
||
let n: usize = obstacle_grid[0].len();
|
||
if obstacle_grid[0][0] == 1 || obstacle_grid[m-1][n-1] == 1 {
|
||
return 0;
|
||
}
|
||
let mut dp = vec![vec![0; n]; m];
|
||
for i in 0..m {
|
||
if obstacle_grid[i][0] == 1 {
|
||
break;
|
||
}
|
||
else { dp[i][0] = 1; }
|
||
}
|
||
for j in 0..n {
|
||
if obstacle_grid[0][j] == 1 {
|
||
break;
|
||
}
|
||
else { dp[0][j] = 1; }
|
||
}
|
||
for i in 1..m {
|
||
for j in 1..n {
|
||
if obstacle_grid[i][j] == 1 {
|
||
continue;
|
||
}
|
||
dp[i][j] = dp[i-1][j] + dp[i][j-1];
|
||
}
|
||
}
|
||
dp[m-1][n-1]
|
||
}
|
||
}
|
||
```
|
||
|
||
空间优化：
|
||
|
||
```rust
|
||
impl Solution {
|
||
pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
|
||
let mut dp = vec![0; obstacle_grid[0].len()];
|
||
for (i, &v) in obstacle_grid[0].iter().enumerate() {
|
||
if v == 0 {
|
||
dp[i] = 1;
|
||
} else {
|
||
break;
|
||
}
|
||
}
|
||
for rows in obstacle_grid.iter().skip(1) {
|
||
for j in 0..rows.len() {
|
||
if rows[j] == 1 {
|
||
dp[j] = 0;
|
||
} else if j != 0 {
|
||
dp[j] += dp[j - 1];
|
||
}
|
||
}
|
||
}
|
||
dp.pop().unwrap()
|
||
}
|
||
}
|
||
```
|
||
|
||
### C
|
||
|
||
```c
|
||
//初始化dp数组
|
||
int **initDP(int m, int n, int** obstacleGrid) {
|
||
int **dp = (int**)malloc(sizeof(int*) * m);
|
||
int i, j;
|
||
//初始化每一行数组
|
||
for(i = 0; i < m; ++i) {
|
||
dp[i] = (int*)malloc(sizeof(int) * n);
|
||
}
|
||
|
||
//先将第一行第一列设为0
|
||
for(i = 0; i < m; ++i) {
|
||
dp[i][0] = 0;
|
||
}
|
||
for(j = 0; j < n; ++j) {
|
||
dp[0][j] = 0;
|
||
}
|
||
|
||
//若碰到障碍，之后的都走不了。退出循环
|
||
for(i = 0; i < m; ++i) {
|
||
if(obstacleGrid[i][0]) {
|
||
break;
|
||
}
|
||
dp[i][0] = 1;
|
||
}
|
||
for(j = 0; j < n; ++j) {
|
||
if(obstacleGrid[0][j])
|
||
break;
|
||
dp[0][j] = 1;
|
||
}
|
||
return dp;
|
||
}
|
||
|
||
int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obstacleGridColSize){
|
||
int m = obstacleGridSize, n = *obstacleGridColSize;
|
||
//初始化dp数组
|
||
int **dp = initDP(m, n, obstacleGrid);
|
||
|
||
int i, j;
|
||
for(i = 1; i < m; ++i) {
|
||
for(j = 1; j < n; ++j) {
|
||
//若当前i,j位置有障碍
|
||
if(obstacleGrid[i][j])
|
||
//路线不同
|
||
dp[i][j] = 0;
|
||
else
|
||
dp[i][j] = dp[i-1][j] + dp[i][j-1];
|
||
}
|
||
}
|
||
//返回最后终点的路径个数
|
||
return dp[m-1][n-1];
|
||
}
|
||
```
|
||
|
||
空间优化版本:
|
||
|
||
```c
|
||
int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obstacleGridColSize){
|
||
int m = obstacleGridSize;
|
||
int n = obstacleGridColSize[0];
|
||
int *dp = (int*)malloc(sizeof(int) * n);
|
||
int i, j;
|
||
|
||
// 初始化dp为第一行起始状态。
|
||
for (j = 0; j < n; ++j) {
|
||
if (obstacleGrid[0][j] == 1)
|
||
dp[j] = 0;
|
||
else if (j == 0)
|
||
dp[j] = 1;
|
||
else
|
||
dp[j] = dp[j - 1];
|
||
}
|
||
|
||
for (i = 1; i < m; ++i) {
|
||
for (j = 0; j < n; ++j) {
|
||
if (obstacleGrid[i][j] == 1)
|
||
dp[j] = 0;
|
||
// 若j为0，dp[j]表示最左边一列，无需改动
|
||
// 此处dp[j],dp[j-1]等同于二维dp中的dp[i-1][j]和dp[i][j-1]
|
||
else if (j != 0)
|
||
dp[j] += dp[j - 1];
|
||
}
|
||
}
|
||
|
||
return dp[n - 1];
|
||
}
|
||
```
|
||
|
||
### Scala
|
||
|
||
```scala
|
||
object Solution {
|
||
import scala.util.control.Breaks._
|
||
def uniquePathsWithObstacles(obstacleGrid: Array[Array[Int]]): Int = {
|
||
var (m, n) = (obstacleGrid.length, obstacleGrid(0).length)
|
||
var dp = Array.ofDim[Int](m, n)
|
||
|
||
// 比如break、continue这些流程控制需要使用breakable
|
||
breakable(
|
||
for (i <- 0 until m) {
|
||
if (obstacleGrid(i)(0) != 1) dp(i)(0) = 1
|
||
else break()
|
||
}
|
||
)
|
||
breakable(
|
||
for (j <- 0 until n) {
|
||
if (obstacleGrid(0)(j) != 1) dp(0)(j) = 1
|
||
else break()
|
||
}
|
||
)
|
||
|
||
for (i <- 1 until m; j <- 1 until n; if obstacleGrid(i)(j) != 1) {
|
||
dp(i)(j) = dp(i - 1)(j) + dp(i)(j - 1)
|
||
}
|
||
|
||
dp(m - 1)(n - 1)
|
||
}
|
||
}
|
||
```
|
||
|
||
<p align="center">
|
||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||
</a>
|