三刷50

Author: diguage

docs/0050-powx-n.adoc

@@ -1,40 +1,39 @@
 [#0050-powx-n]
 = 50. Pow(x, n)
 
-{leetcode}/problems/powx-n/[LeetCode - Pow(x, n)^]
+https://leetcode.cn/problems/powx-n/[LeetCode - 50. Pow(x, n) ^]
 
-Implement http://www.cplusplus.com/reference/valarray/pow/[pow(_x_, _n_)^], which calculates _x_ raised to the power _n_ (x^n^).
+实现 https://www.cplusplus.com/reference/valarray/pow/[pow(x, n)] ，即计算 `x` 的整数 `n` 次幂函数（即，`x^n^`）。
 
-*Example 1:*
+*示例 1：*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* 2.00000, 10
-*Output:* 1024.00000
-----
+....
+输入：x = 2.00000, n = 10
+输出：1024.00000
+....
 
-*Example 2:*
+*示例 2：*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* 2.10000, 3
-*Output:* 9.26100
-----
+....
+输入：x = 2.10000, n = 3
+输出：9.26100
+....
 
-*Example 3:*
-
-[subs="verbatim,quotes,macros"]
-----
-*Input:* 2.00000, -2
-*Output:* 0.25000
-*Explanation:* 2^-2^ = 1/2^2^ = 1/4 = 0.25
-----
+*示例 3：*
 
-*Note:*
+....
+输入：x = 2.00000, n = -2
+输出：0.25000
+解释：2-2 = 1/22 = 1/4 = 0.25
+....
 
+*提示：*
 
-* -100.0 < _x_ < 100.0
-* _n_ is a 32-bit signed integer, within the range [-2^31^, 2^31 ^- 1]
+* `-100.0 < x < 100.0`
+* `-2^31^ \<= n \<= 2^31^-1`
+* `n` 是一个整数
+* 要么 `x` 不为零，要么 `n > 0` 。
+* `-10^4^ \<= x^n^ \<= 10^4^`
 
 == 思路分析
 
@@ -70,11 +69,20 @@ include::{sourcedir}/_0050_PowXN.java[tag=answer]
 include::{sourcedir}/_0050_PowXN_2.java[tag=answer]
 ----
 --
+
+三刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0050_PowXN_3.java[tag=answer]
+----
+--
 ====
 
 == 参考资料
 
-. https://leetcode.cn/problems/powx-n/solutions/238559/powx-n-by-leetcode-solution/?envType=study-plan-v2&envId=selected-coding-interview[50. Pow(x, n) - 官方题解^]
-. https://leetcode.cn/problems/powx-n/solutions/241471/50-powx-n-kuai-su-mi-qing-xi-tu-jie-by-jyd/?envType=study-plan-v2&envId=selected-coding-interview[50. Pow(x, n) - 快速幂，清晰图解^]
+. https://leetcode.cn/problems/powx-n/solutions/238559/powx-n-by-leetcode-solution/[50. Pow(x, n) - 官方题解^]
+. https://leetcode.cn/problems/powx-n/solutions/241471/50-powx-n-kuai-su-mi-qing-xi-tu-jie-by-jyd/[50. Pow(x, n) - 快速幂，清晰图解^]
 
 

logbook/202503.adoc

@@ -185,6 +185,11 @@ endif::[]
 |{doc_base_url}/0088-merge-sorted-array.adoc[题解]
 |✅ `num1` 后面空着，则从后向前合并。
 
+|{counter:codes2503}
+|{leetcode_base_url}/powx-n/[50. Pow(x, n)^]
+|{doc_base_url}/0050-powx-n.adoc[题解]
+|✅ 递归很简单，抽空再思考一下非递归形式。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

src/main/java/com/diguage/algo/leetcode/_0050_PowXN_3.java

@@ -0,0 +1,28 @@
+package com.diguage.algo.leetcode;
+
+public class _0050_PowXN_3 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-04-07 14:26:30
+   */
+  public double myPow(double x, long n) {
+    if (x == 0) {
+      return 0;
+    }
+    if (n == 0) {
+      return 1;
+    }
+    boolean negative = n < 0;
+    n = Math.abs(n);
+    double result = 1;
+    double bin = myPow(x, n / 2);
+    if (n % 2 == 1) {
+      result = x * bin * bin;
+    } else {
+      result = bin * bin;
+    }
+    return negative ? 1 / result : result;
+  }
+  // end::answer[]
+}