From 75ad4dda7ce6e4cae5f0e43b8ba6a5dcaf45a99a Mon Sep 17 00:00:00 2001
From: CyC2018 <zhengyc101@163.com>
Date: Wed, 28 Oct 2026 13:45:03 +0800
Subject: [PATCH] =?UTF-8?q?Update=2014.=20=E5=89=AA=E7=BB=B3=E5=AD=90.md?=
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题目链接改为力扣中的对应题目，题目描述和代码按该题的接口调整。
---
 notes/14. 剪绳子.md | 24 ++++++++++++------------
 1 file changed, 12 insertions(+), 12 deletions(-)

diff --git a/notes/14. 剪绳子.md b/notes/14. 剪绳子.md
index 9108b635..2996240e 100644
--- a/notes/14. 剪绳子.md	
+++ b/notes/14. 剪绳子.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/57d85990ba5b440ab888fc72b0751bf8?tpId=13&tqId=33257&tab=answerKey&from=cyc_github)
+[Leetcode](https://leetcode.cn/problems/jian-sheng-zi-lcof/)
 
 ## 题目描述
 
@@ -37,17 +37,17 @@ return 36 (10 = 3 + 3 + 4)
 继续拆成更大的绳子可以发现都比拆成 2 和 3 的效果更差，因此我们只考虑将绳子拆成 2 和 3，并且优先拆成 3，当拆到绳子长度 n 等于 4 时，也就是出现 3+1，此时只能拆成 2+2。
 
 ```java
-public int cutRope(int n) {
-    if (n < 2)
+public int cuttingBamboo(int bamboo_len) {
+    if (bamboo_len < 2)
         return 0;
-    if (n == 2)
+    if (bamboo_len == 2)
         return 1;
-    if (n == 3)
+    if (bamboo_len == 3)
         return 2;
-    int timesOf3 = n / 3;
-    if (n - timesOf3 * 3 == 1)
+    int timesOf3 = bamboo_len / 3;
+    if (bamboo_len - timesOf3 * 3 == 1)
         timesOf3--;
-    int timesOf2 = (n - timesOf3 * 3) / 2;
+    int timesOf2 = (bamboo_len - timesOf3 * 3) / 2;
     return (int) (Math.pow(3, timesOf3)) * (int) (Math.pow(2, timesOf2));
 }
 ```
@@ -55,13 +55,13 @@ public int cutRope(int n) {
 ### 动态规划
 
 ```java
-public int cutRope(int n) {
-    int[] dp = new int[n + 1];
+public int cuttingBamboo(int bamboo_len) {
+    int[] dp = new int[bamboo_len + 1];
     dp[1] = 1;
-    for (int i = 2; i <= n; i++)
+    for (int i = 2; i <= bamboo_len; i++)
         for (int j = 1; j < i; j++)
             dp[i] = Math.max(dp[i], Math.max(j * (i - j), dp[j] * (i - j)));
-    return dp[n];
+    return dp[bamboo_len];
 }
 ```