一刷746

Author: diguage

README.adoc

@@ -5247,14 +5247,14 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|{doc_base_url}/0745-prefix-and-suffix-search.adoc[题解]
 //|Hard
 //|
-//
-//|{counter:codes}
-//|{leetcode_base_url}/min-cost-climbing-stairs/[746. Min Cost Climbing Stairs^]
-//|{source_base_url}/_0746_MinCostClimbingStairs.java[Java]
-//|{doc_base_url}/0746-min-cost-climbing-stairs.adoc[题解]
-//|Easy
-//|
-//
+
+|{counter:codes}
+|{leetcode_base_url}/min-cost-climbing-stairs/[746. Min Cost Climbing Stairs^]
+|{source_base_url}/_0746_MinCostClimbingStairs.java[Java]
+|{doc_base_url}/0746-min-cost-climbing-stairs.adoc[题解]
+|Easy
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/largest-number-at-least-twice-of-others/[747. Largest Number At Least Twice of Others^]
 //|{source_base_url}/_0747_LargestNumberAtLeastTwiceOfOthers.java[Java]

docs/0736-parse-lisp-expression.adoc

@@ -68,11 +68,33 @@ of the final x in the add-expression.  That final x will equal 2.
 . The length of `expression` is at most 2000.  (It is also non-empty, as that would not be a legal expression.)</li>
 . The answer and all intermediate calculations of that answer are guaranteed to fit in a 32-bit integer.</li>
 
+== 思路分析
+
+image::images/0746-01.jpg[{image_attr}]
 
 
 [[src-0736]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0736_ParseLispExpression.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0078_Subsets_2.java[tag=answer]
+// ----
+// --
+====
+
+== 参考资料
+
 

docs/0746-min-cost-climbing-stairs.adoc

@@ -38,10 +38,36 @@ Once you pay the cost, you can either climb one or two steps. You need to find m
 
 
 
+== 思路分析
 
-[[src-0746]]
+image::images/0746-01.jpg[{image_attr}]
+
+
+[[src-0736]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0746_MinCostClimbingStairs.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0078_Subsets_2.java[tag=answer]
+// ----
+// --
+====
+
+== 参考资料
+
+. https://leetcode.cn/problems/min-cost-climbing-stairs/solutions/528955/shi-yong-zui-xiao-hua-fei-pa-lou-ti-by-l-ncf8/[746. 使用最小花费爬楼梯 - 官方题解^]
+. https://leetcode.cn/problems/min-cost-climbing-stairs/solutions/177077/yi-bu-yi-bu-tui-dao-dong-tai-gui-hua-de-duo-chong-/[746. 使用最小花费爬楼梯 - 一步一步推导动态规划的多种解法^]
+. https://leetcode.cn/problems/min-cost-climbing-stairs/solutions/856958/dai-ma-sui-xiang-lu-dong-tai-gui-hua-wu-abbwz/[746. 使用最小花费爬楼梯 - 「代码随想录」带你学透动态规划！^]
 

docs/images/0746-01.jpg

@@ -1566,7 +1566,7 @@ include::0744-find-smallest-letter-greater-than-target.adoc[leveloffset=+1]
 
 // include::0745-prefix-and-suffix-search.adoc[leveloffset=+1]
 
-// include::0746-min-cost-climbing-stairs.adoc[leveloffset=+1]
+include::0746-min-cost-climbing-stairs.adoc[leveloffset=+1]
 
 // include::0747-largest-number-at-least-twice-of-others.adoc[leveloffset=+1]
 

docs/index.adoc

@@ -449,6 +449,11 @@
 |{doc_base_url}/0037-sudoku-solver.adoc[题解]
 |回溯
 
+|{counter:codes}
+|{leetcode_base_url}/min-cost-climbing-stairs/[746. Min Cost Climbing Stairs^]
+|{doc_base_url}/0746-min-cost-climbing-stairs.adoc[题解]
+|动态规划
+
 
 |===
 

logbook/202406.adoc

@@ -23,7 +23,7 @@
     <maven-javadoc-plugin.version>3.8.0</maven-javadoc-plugin.version>
     <fizzed-watcher-maven-plugin.version>3.7.0</fizzed-watcher-maven-plugin.version>
 
-    <asciidoctorj.version>3.0.0-beta.1</asciidoctorj.version>
+    <asciidoctorj.version>3.0.0</asciidoctorj.version>
     <asciidoctorj-pdf.version>2.3.18</asciidoctorj-pdf.version>
     <asciidoctorj-epub3.version>2.1.3</asciidoctorj-epub3.version>
     <asciidoctorj-diagram.version>2.3.1</asciidoctorj-diagram.version>

pom.xml

@@ -0,0 +1,25 @@
+package com.diguage.algo.leetcode;
+
+public class _0746_MinCostClimbingStairs {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-09-11 20:23:46
+   */
+  public int minCostClimbingStairs(int[] cost) {
+    // dp[i] 表示到底第 i 阶时，所需最少费用
+    int[] dp = new int[cost.length + 1];
+    // 0 和 1 可以直接站上，所以不需要付费
+    dp[0] = 0;
+    dp[1] = 0;
+    for (int i = 2; i < dp.length; i++) {
+      // 第 i 阶，有两种走法
+      // 1. 从第 i-1 阶走一步上去，花费： dp[i-1] 走到 i-1 的费用 + cost[i-1]（向上的费用）
+      // 2. 从第 i-2 阶走两步上去，花肥： dp[i-2] 走到 i-2 的费用 + cost[i-2]（向上的费用）
+      dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
+    }
+    return dp[dp.length - 1];
+  }
+  // end::answer[]
+}