三刷70

Author: diguage

docs/0070-climbing-stairs.adoc

@@ -1,36 +1,37 @@
 [#0070-climbing-stairs]
-= 70. Climbing Stairs
+= 70. 爬楼梯
 
-{leetcode}/problems/climbing-stairs/[LeetCode - Climbing Stairs^]
+https://leetcode.cn/problems/climbing-stairs/[LeetCode - 70. 爬楼梯^]
 
-You are climbing a stair case. It takes _n_ steps to reach to the top.
+假设你正在爬楼梯。需要 `n` 阶你才能到达楼顶。
 
-Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
+每次你可以爬 `1` 或 `2` 个台阶。你有多少种不同的方法可以爬到楼顶呢？
 
-*Note:* Given _n_ will be a positive integer.
+*示例 1：*
 
-*Example 1:*
+....
+输入：n = 2
+输出：2
+解释：有两种方法可以爬到楼顶。
+1. 1 阶 + 1 阶
+2. 2 阶
+....
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* 2
-*Output:* 2
-*Explanation:* There are two ways to climb to the top.
-1. 1 step + 1 step
-2. 2 steps
-----
+*示例 2：*
 
-*Example 2:*
+....
+输入：n = 3
+输出：3
+解释：有三种方法可以爬到楼顶。
+1. 1 阶 + 1 阶 + 1 阶
+2. 1 阶 + 2 阶
+3. 2 阶 + 1 阶
+....
+
+*提示：*
+
+* `+1 <= n <= 45+`
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* 3
-*Output:* 3
-*Explanation:* There are three ways to climb to the top.
-1. 1 step + 1 step + 1 step
-2. 1 step + 2 steps
-3. 2 steps + 1 step
-----
 
 == 思路分析
 
@@ -79,13 +80,22 @@ include::{sourcedir}/_0070_ClimbingStairs.java[tag=answer]
 include::{sourcedir}/_0070_ClimbingStairs_2.java[tag=answer]
 ----
 --
-====
-
-== 参考资料
 
-. https://leetcode.cn/problems/climbing-stairs/solutions/2361764/70-pa-lou-ti-dong-tai-gui-hua-qing-xi-tu-ruwa/?envType=study-plan-v2&envId=selected-coding-interview[70. 爬楼梯 - 动态规划，清晰图解^]
-. https://leetcode.cn/problems/climbing-stairs/solutions/286022/pa-lou-ti-by-leetcode-solution/?envType=study-plan-v2&envId=selected-coding-interview[70. 爬楼梯 - 官方题解^]
-. https://leetcode.cn/problems/climbing-stairs/solutions/2560716/jiao-ni-yi-bu-bu-si-kao-dong-tai-gui-hua-7zm1/?envType=study-plan-v2&envId=selected-coding-interview[70. 爬楼梯 - 教你一步步思考动态规划：从记忆化搜索到递推^]
+三刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0070_ClimbingStairs_3.java[tag=answer]
+----
+--
+====
 
 
+== 参考资料
 
+. https://leetcode.cn/problems/climbing-stairs/solutions/2560716/jiao-ni-yi-bu-bu-si-kao-dong-tai-gui-hua-7zm1/[70. 爬楼梯 - 教你一步步思考动态规划：从记忆化搜索到递推^]
+. https://leetcode.cn/problems/climbing-stairs/solutions/270926/cong-zhi-jue-si-wei-fen-xi-dong-tai-gui-hua-si-lu-/[70. 爬楼梯 - 「手画图解」详解爬楼梯问题 | 从递归，讲到动态规划^]
+. https://leetcode.cn/problems/climbing-stairs/solutions/854668/dai-ma-sui-xiang-lu-dong-tai-gui-hua-jin-y1hw/[70. 爬楼梯 - 「代码随想录」动态规划精讲！^]
+. https://leetcode.cn/problems/climbing-stairs/solutions/2361764/70-pa-lou-ti-dong-tai-gui-hua-qing-xi-tu-ruwa/[70. 爬楼梯 - 动态规划，清晰图解^]
+. https://leetcode.cn/problems/climbing-stairs/solutions/286022/pa-lou-ti-by-leetcode-solution/[70. 爬楼梯 - 官方题解^]

logbook/202503.adoc

@@ -1954,6 +1954,11 @@ endif::[]
 |{doc_base_url}/0072-edit-distance.adoc[题解]
 |❌ 动态规划。计算 `dp[c][r]` 时，从相邻三个元素获取递推的含义还要好好思考一下，“增”、“删”和“改”的含义要理解清楚。
 
+|{counter:codes2503}
+|{leetcode_base_url}/climbing-stairs/[70. 爬楼梯^]
+|{doc_base_url}/0070-climbing-stairs.adoc[题解]
+|✅ 动态规划。也就是斐波那契数列，没想到这竟然是最简单的动态规划题目。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

src/main/java/com/diguage/algo/leetcode/_0070_ClimbingStairs_3.java

@@ -0,0 +1,25 @@
+package com.diguage.algo.leetcode;
+
+public class _0070_ClimbingStairs_3 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-11-21 23:18:37
+   */
+  public int climbStairs(int n) {
+    if (n == 1) {
+      return 1;
+    }
+    if (n == 2) {
+      return 2;
+    }
+    int a = 1, b = 2;
+    for (int i = 3; i <= n; i++) {
+      int sum = a + b;
+      a = b;
+      b = sum;
+    }
+    return b;
+  }
+  // end::answer[]
+}