三刷343

Author: diguage

docs/0343-integer-break.adoc

@@ -54,6 +54,15 @@ include::{sourcedir}/_0343_IntegerBreak.java[tag=answer]
 include::{sourcedir}/_0343_IntegerBreak_2.java[tag=answer]
 ----
 --
+
+三刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0343_IntegerBreak_3.java[tag=answer]
+----
+--
 ====
 
 == 参考资料

logbook/202406.adoc

@@ -595,6 +595,10 @@
 |{doc_base_url}/0231-power-of-two.adoc[题解]
 |位运算， `n & (n - 1)` 或约数
 
+|{counter:codes}
+|{leetcode_base_url}/integer-break/[343. Integer Break^]
+|{doc_base_url}/0343-integer-break.adoc[题解]
+|动态规划，数学分析也可以看一下：怎么样计算自然底数 `e` 的值？
 
 |===
 

src/main/java/com/diguage/algo/leetcode/_0343_IntegerBreak_3.java

@@ -0,0 +1,21 @@
+package com.diguage.algo.leetcode;
+
+public class _0343_IntegerBreak_3 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-09-15 20:47:03
+   */
+  public int integerBreak(int n) {
+    int[] dp = new int[n + 1];
+    for (int i = 2; i <= n; i++) {
+      int max = -1;
+      for (int j = 1; j < i / 2 + 1; j++) {
+        max = Math.max(Math.max(max, j * (i - j)), j * dp[i - j]);
+      }
+      dp[i] = max;
+    }
+    return dp[n];
+  }
+  // end::answer[]
+}