二刷62

Author: diguage

docs/0000-26-dynamic-programming.adoc

@@ -7,6 +7,14 @@ image::images/dynamic-programming-01.jpeg[{image_attr}]
 
 :sectnums:
 
+动态规划的五部曲：
+
+. 确定 dp 数组（dp table）以及下标的含义
+. 确定递推公式
+. dp 数组如何初始化
+. 确定遍历顺序
+. 举例推导 dp 数组
+
 include::0000-26-dp-1-0-1-knapsack.adoc[leveloffset=+1]
 
 include::0000-26-dp-2-unbounded-knapsack.adoc[leveloffset=+1]

docs/0062-unique-paths.adoc

@@ -38,10 +38,34 @@ From the top-left corner, there are a total of 3 ways to reach the bottom-right
 *Output:* 28
 ----
 
+== 思路分析
 
 [[src-0062]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0062_UniquePaths.java[tag=answer]
 ----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0062_UniquePaths_2.java[tag=answer]
+----
+--
+====
+
+== 参考资料
+
+. https://leetcode.cn/problems/unique-paths/solutions/5772/dong-tai-gui-hua-by-powcai-2/[62. 不同路径 - 力扣（LeetCode）^]
+. https://leetcode.cn/problems/unique-paths/solutions/514311/bu-tong-lu-jing-by-leetcode-solution-hzjf/[62. 不同路径 - 官方题解^]
+
+
 

logbook/202406.adoc

@@ -454,6 +454,10 @@
 |{doc_base_url}/0746-min-cost-climbing-stairs.adoc[题解]
 |动态规划
 
+|{counter:codes}
+|{leetcode_base_url}/unique-paths/[62. Unique Paths^]
+|{doc_base_url}/0062-unique-paths.adoc[题解]
+|动态规划
 
 |===
 

src/main/java/com/diguage/algo/leetcode/_0062_UniquePaths.java

@@ -39,6 +39,11 @@
  */
 public class _0062_UniquePaths {
   // tag::answer[]
+
+    /**
+     * @author D瓜哥 · https://www.diguage.com
+     * @since 2019-10-26 22:56
+     */
     public int uniquePaths(int m, int n) {
         if (m == 0 || n == 0) {
             return 0;

src/main/java/com/diguage/algo/leetcode/_0062_UniquePaths_2.java

@@ -0,0 +1,31 @@
+package com.diguage.algo.leetcode;
+
+public class _0062_UniquePaths_2 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-09-12 15:16:31
+   */
+  public int uniquePaths(int m, int n) {
+    // 1. dp 表示走到某个节点一共有多少路径
+    int[][] dp = new int[m][n];
+    // 3. 第一行只有一种走法
+    for (int i = 0; i < m; i++) {
+      dp[i][0] = 1;
+    }
+    // 3. 第一列也只有一种走法。
+    for (int i = 0; i < n; i++) {
+      dp[0][i] = 1;
+    }
+    // 4. 遍历顺序：从左上逐步向右下遍历
+    for (int i = 1; i < m; i++) {
+      for (int j = 1; j < n; j++) {
+        // 2. 由于只能从上或者从左走过来，
+        //    那么路径数量就是两个节点相加
+        dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
+      }
+    }
+    return dp[m - 1][n - 1];
+  }
+  // end::answer[]
+}