二刷980

Author: diguage

docs/0980-unique-paths-iii.adoc

@@ -1,7 +1,7 @@
 [#0980-unique-paths-iii]
 = 980. 不同路径 III
 
-https://leetcode.cn/problems/unique-paths-iii/[LeetCode - 980. 不同路径 III ^]
+https://leetcode.cn/problems/unique-paths-iii/[LeetCode - 980. 不同路径 III^]
 
 在二维网格 `grid` 上，有 4 种类型的方格：
 
@@ -16,8 +16,14 @@ https://leetcode.cn/problems/unique-paths-iii/[LeetCode - 980. 不同路径 III
 
 *示例 1：*
 
+image::images/0980-01.jpg[{image_attr}]
+
 ....
-输入：[[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
+输入：[
+       [1,0,0,0],
+       [0,0,0,0],
+       [0,0,2,-1]
+     ]
 输出：2
 解释：我们有以下两条路径：
 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
@@ -26,8 +32,14 @@ https://leetcode.cn/problems/unique-paths-iii/[LeetCode - 980. 不同路径 III
 
 *示例 2：*
 
+image::images/0980-02.jpg[{image_attr}]
+
 ....
-输入：[[1,0,0,0],[0,0,0,0],[0,0,0,2]]
+输入：[
+       [1,0,0,0],
+       [0,0,0,0],
+       [0,0,0,2]
+     ]
 输出：4
 解释：我们有以下四条路径：
 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
@@ -38,8 +50,13 @@ https://leetcode.cn/problems/unique-paths-iii/[LeetCode - 980. 不同路径 III
 
 *示例 3：*
 
+image::images/0980-03.jpg[{image_attr}]
+
 ....
-输入：[[0,1],[2,0]]
+输入：[
+       [0,1],
+       [2,0]
+     ]
 输出：0
 解释：
 没有一条路能完全穿过每一个空的方格一次。
@@ -66,17 +83,19 @@ include::{sourcedir}/_0980_UniquePathsIii.java[tag=answer]
 ----
 --
 
-// 二刷::
-// +
-// --
-// [{java_src_attr}]
-// ----
-// include::{sourcedir}/_0980_UniquePathsIii_2.java[tag=answer]
-// ----
-// --
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0980_UniquePathsIii_2.java[tag=answer]
+----
+--
 ====
 
 
 == 参考资料
 
-
+. https://leetcode.cn/problems/unique-paths-iii/solutions/2372252/liang-chong-fang-fa-hui-su-zhuang-tai-ya-26py/[980. 不同路径 III - 两种方法：回溯/状态压缩+记忆化搜索^]
+. https://leetcode.cn/problems/unique-paths-iii/solutions/317829/dfs-hui-su-shuang-bai-by-quantum-10/[980. 不同路径 III - dfs + 回溯 (双百)^]
+. https://leetcode.cn/problems/unique-paths-iii/solutions/2365866/bu-tong-lu-jing-iii-by-leetcode-solution-cndw/[980. 不同路径 III - 官方题解^]

docs/images/0980-01.jpg

@@ -88,7 +88,7 @@ endif::[]
 |{counter:codes2503e}
 |{leetcode_base_url}/unique-paths-iii/[980. 不同路径 III^]
 |{doc_base_url}/0980-unique-paths-iii.adoc[题解]
-|❌
+|❌ ✅
 
 |{counter:codes2503e}
 |{leetcode_base_url}/minimum-falling-path-sum-ii/[1289. 下降路径最小和  II^]
@@ -2665,6 +2665,11 @@ endif::[]
 |{doc_base_url}/0740-delete-and-earn.adoc[题解]
 |❌ 动态规划。最开始以为理解所有连续数字的和，后来发现理解错了。后来理解正确后，选定一个，删除上下紧挨的两个数，然后再选。这就是打家劫舍的思路了。
 
+|{counter:codes2503}
+|{leetcode_base_url}/unique-paths-iii/[980. 不同路径 III^]
+|{doc_base_url}/0980-unique-paths-iii.adoc[题解]
+|✅ 回溯。找到起点，然后向四面八方扩散。
+
 |===
 
 

docs/images/0980-02.jpg

@@ -0,0 +1,68 @@
+package com.diguage.algo.leetcode;
+
+public class _0980_UniquePathsIii_2 {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-12-21 16:39:19
+   */
+  private int result = 0;
+
+  public int uniquePathsIII(int[][] grid) {
+    int[] start = new int[2];
+    int[] end = new int[2];
+    int step = 0;
+    for (int c = 0; c < grid.length; c++) {
+      for (int r = 0; r < grid[c].length; r++) {
+        if (grid[c][r] == 1) {
+          start[0] = c;
+          start[1] = r;
+        } else if (grid[c][r] == 2) {
+          end[0] = c;
+          end[1] = r;
+        } else if (grid[c][r] == 0) {
+          step++;
+        }
+      }
+    }
+    backtrack(grid, start, end, step + 1);
+    return result;
+  }
+
+  private void backtrack(int[][] grid, int[] start, int[] end, int step) {
+    int column = start[0];
+    int row = start[1];
+    if (step == 0 && column == end[0] && row == end[1]) {
+      result++;
+      return;
+    }
+    if (column < 0 || column >= grid.length
+      || row < 0 || row >= grid[column].length
+      || step < 0
+      || (grid[column][row] != 1 && grid[column][row] != 0)) {
+      return;
+    }
+    int origin = grid[column][row];
+    int nextStep = step - 1;
+    grid[column][row] = Integer.MAX_VALUE;
+    // 上
+    backtrack(grid, new int[]{column - 1, row}, end, nextStep);
+    // 下
+    backtrack(grid, new int[]{column + 1, row}, end, nextStep);
+    // 左
+    backtrack(grid, new int[]{column, row - 1}, end, nextStep);
+    // 右
+    backtrack(grid, new int[]{column, row + 1}, end, nextStep);
+
+    grid[column][row] = origin;
+  }
+  // end::answer[]
+
+  static void main() {
+    new _0980_UniquePathsIii_2().uniquePathsIII(new int[][]{
+      {1, 0, 0, 0},
+      {0, 0, 0, 0},
+      {0, 0, 2, -1}});
+  }
+}