二刷63

Author: diguage

docs/0063-unique-paths-ii.adoc

@@ -33,10 +33,34 @@ There are two ways to reach the bottom-right corner:
 2. Down -> Down -> Right -> Right
 ----
 
+== 思路分析
+
+image::images/0063-01.png[{image_attr}]
 
 [[src-0063]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0063_UniquePathsII.java[tag=answer]
 ----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0063_UniquePathsII_2.java[tag=answer]
+----
+--
+====
+
+== 参考资料
+
+. https://leetcode.cn/problems/unique-paths-ii/solutions/316968/bu-tong-lu-jing-ii-by-leetcode-solution-2/[63. 不同路径 II - 官方题解^]
+. https://leetcode.cn/problems/unique-paths-ii/solutions/2732049/javapython3cdong-tai-gui-hua-kong-jian-y-nyjy/[63. 不同路径 II - 动态规划 + 空间优化^]
 

docs/images/0063-01.png

@@ -459,6 +459,11 @@
 |{doc_base_url}/0062-unique-paths.adoc[题解]
 |动态规划
 
+|{counter:codes}
+|{leetcode_base_url}/unique-paths-ii/[63. Unique Paths II^]
+|{doc_base_url}/0063-unique-paths-ii.adoc[题解]
+|动态规划
+
 |===
 
 截止目前，本轮练习一共完成 {codes} 道题。

logbook/202406.adoc

@@ -43,6 +43,9 @@ public class _0063_UniquePathsII {
      * Runtime: 0 ms, faster than 100.00% of Java online submissions for Unique Paths II.
      * <p>
      * Memory Usage: 40.5 MB, less than 33.84% of Java online submissions for Unique Paths II.
+     *
+     * @author D瓜哥 · https://www.diguage.com
+     * @since 2019-10-26 23:50
      */
     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
         if (Objects.isNull(obstacleGrid)

src/main/java/com/diguage/algo/leetcode/_0063_UniquePathsII.java

@@ -0,0 +1,66 @@
+package com.diguage.algo.leetcode;
+
+import static com.diguage.util.Printers.printMatrix;
+
+public class _0063_UniquePathsII_2 {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2019-10-26 23:50
+   */
+  public int uniquePathsWithObstacles(int[][] obstacleGrid) {
+    int row = obstacleGrid.length;
+    int column = obstacleGrid[0].length;
+    // 1. obstacleGrid 表示走到某个节点一共有多少路径
+    //    将障碍设置为-1，便于和 0 区分
+    for (int i = 0; i < row; i++) {
+      for (int j = 0; j < column; j++) {
+        if (obstacleGrid[i][j] == 1) {
+          obstacleGrid[i][j] = -1;
+        }
+      }
+    }
+    // 3. 第一行也只有一种走法，如果有障碍，则后面的路径为0
+    for (int i = 0; i < column; i++) {
+      if (obstacleGrid[0][i] == -1) {
+        break;
+      }
+      obstacleGrid[0][i] = 1;
+    }
+    // 3. 第一列只有一种走法，如果有障碍，则后面的路径为0
+    for (int i = 0; i < row; i++) {
+      if (obstacleGrid[i][0] == -1) {
+        break;
+      }
+      obstacleGrid[i][0] = 1;
+    }
+    printMatrix(obstacleGrid);
+    // 4. 确定遍历顺序：从左上向右下遍历
+    for (int i = 1; i < row; i++) {
+      for (int j = 1; j < column; j++) {
+        if (obstacleGrid[i][j] == -1) {
+          continue;
+        }
+        // 2. 确定递推公式
+        int mi = 0;
+        if (obstacleGrid[i - 1][j] != -1) {
+          mi = obstacleGrid[i - 1][j];
+        }
+        int ni = 0;
+        if (obstacleGrid[i][j - 1] != -1) {
+          ni = obstacleGrid[i][j - 1];
+        }
+        obstacleGrid[i][j] = mi + ni;
+        printMatrix(obstacleGrid);
+      }
+    }
+    return obstacleGrid[row - 1][column - 1] == -1 ? 0 : obstacleGrid[row - 1][column - 1];
+  }
+
+  // end::answer[]
+  public static void main(String[] args) {
+    int[][] obstacleGrid = {{0, 0, 0}, {0, 1, 0}, {0, 0, 0}};
+    new _0063_UniquePathsII_2().uniquePathsWithObstacles(obstacleGrid);
+  }
+}