2019-11-02 12:07:41 +08:00
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# 14. 剪绳子
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[Leetcode](https://leetcode.com/problems/integer-break/description/)
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## 题目描述
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把一根绳子剪成多段，并且使得每段的长度乘积最大。
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```html
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n = 2
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return 1 (2 = 1 + 1)
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n = 10
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return 36 (10 = 3 + 3 + 4)
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```
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## 解题思路
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### 贪心
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尽可能多剪长度为 3 的绳子，并且不允许有长度为 1 的绳子出现。如果出现了，就从已经切好长度为 3 的绳子中拿出一段与长度为 1 的绳子重新组合，把它们切成两段长度为 2 的绳子。
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证明：当 n >= 5 时，3(n - 3) - n = 2n - 9 > 0，且 2(n - 2) - n = n - 4 > 0。因此在 n >= 5 的情况下，将绳子剪成一段为 2 或者 3，得到的乘积会更大。又因为 3(n - 3) - 2(n - 2) = n - 5 >= 0，所以剪成一段长度为 3 比长度为 2 得到的乘积更大。
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```java
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public int integerBreak(int n) {
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if (n < 2)
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return 0;
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if (n == 2)
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return 1;
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if (n == 3)
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return 2;
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int timesOf3 = n / 3;
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if (n - timesOf3 * 3 == 1)
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timesOf3--;
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int timesOf2 = (n - timesOf3 * 3) / 2;
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return (int) (Math.pow(3, timesOf3)) * (int) (Math.pow(2, timesOf2));
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}
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```
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### 动态规划
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```java
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public int integerBreak(int n) {
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int[] dp = new int[n + 1];
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dp[1] = 1;
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for (int i = 2; i <= n; i++)
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for (int j = 1; j < i; j++)
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dp[i] = Math.max(dp[i], Math.max(j * (i - j), dp[j] * (i - j)));
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return dp[n];
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}
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```
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2019-11-02 17:33:10 +08:00
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
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