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‎README.md

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| 49 | [Group Anagrams][0049] | Hash Table, String |
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| 50 | [Pow(x, n)][0050] | Math, Binary Search |
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| 56 | [Merge Intervals][0056] | Array, Sort |
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| 63 | [不同路径 II(Unique Paths II)][0063] | 数组、动态规划 |
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| 209 | [长度最小的子数组(Minimum Size Subarray Sum)][0209] | 数组、双指针、二分查找 |
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| 215 | [数组中的第K个最大元素(Kth Largest Element in an Array)][0215] | 堆、分治算法 |
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| 554 | [Brick Wall][0554] | Hash Table |
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[0049]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/0049/README.md
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[0050]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/0050/README.md
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[0056]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/0056/README.md
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[0063]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/0063/README.md
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[0209]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/0209/README.md
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[0215]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/0215/README.md
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[0554]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/0554/README.md

‎note/0063/README.md

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# [不同路径 II(Unique Paths II)][title]
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## 题目描述
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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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![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/22/robot_maze.png)
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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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**说明：**_m_ 和 _n_ 的值均不超过 100。
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**示例 1:**
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```
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输入:
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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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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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```
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**标签：** 数组、动态规划
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## 思路
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做过爬楼梯的应该很快就能想到这是一道很典型的动态规划题目，
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我们令 `dp[i][j]` 表示走到格子 `(i, j)` 的路径数，
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那么当 `(i, j)` 没障碍物时，`dp[i][j] = 0`；
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那么当 `(i, j)` 有障碍物时，`dp[i][j] = dp[i - 1][j] + dp[i][j - 1]`；
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其初始态第 1 列（行）的格子只有从其上（左）边格子走过去这一种走法，因此初始化 `dp[i][0]`（`dp[0][j]`）值为 1，且遇到障碍物时后面值都为 0；
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有了这些条件，我相信你肯定可以写出代码来了，具体如下所示：
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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, n = obstacleGrid[0].length;
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int[][] dp = new int[m][n];
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// 其初始态第 1 列（行）的格子只有从其上（左）边格子走过去这一种走法，
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// 因此初始化 dp[i][0]（dp[0][j]）值为 1，且遇到障碍物时后面值都为 0；
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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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if (obstacleGrid[i][j] == 0) {
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// 当 (i, j) 有障碍物时，dp[i][j] = dp[i - 1][j] + dp[i][j - 1]；
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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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return dp[m - 1][n - 1];
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}
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}
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```
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## 结语
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如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-java-leetcode][ajl]
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[title]: https://leetcode-cn.com/problems/unique-paths-ii
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[ajl]: https://github.com/Blankj/awesome-java-leetcode

‎src/com/blankj/medium/_0057/Solution.java renamed to ‎src/com/blankj/hard/_0057/Solution.java

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package com.blankj.medium._057;
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package com.blankj.hard._0057;
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import com.blankj.structure.Interval;
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‎src/com/blankj/medium/_0067/Solution.java

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package com.blankj.medium._0067;
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/**
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* <pre>
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* author: Blankj
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* blog : http://blankj.com
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* time : 2020/07/07
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* desc :
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* </pre>
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*/
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public class Solution {
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public int uniquePathsWithObstacles(int[][] obstacleGrid) {
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int m = obstacleGrid.length, n = obstacleGrid[0].length;
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int[][] dp = new int[m][n];
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// 其初始态第 1 列（行）的格子只有从其上（左）边格子走过去这一种走法，
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// 因此初始化 dp[i][0]（dp[0][j]）值为 1，且遇到障碍物时后面值都为 0；
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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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if (obstacleGrid[i][j] == 0) {
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// 当 (i, j) 有障碍物时，dp[i][j] = dp[i - 1][j] + dp[i][j - 1]；
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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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return dp[m - 1][n - 1];
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}
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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, 0, 0}, {0, 1, 0}, {0, 0, 0}};
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System.out.println(solution.uniquePathsWithObstacles(obstacleGrid));
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}
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}

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