JZ-067-Cut the rope

cut the rope

Topic description

Give you a rope of length n, please cut the rope into m sections of integer length (m, n are integers, n>1 and m>1, m<=n), the length of each rope is recorded as k [1],...,k[m].

• What is the largest possible product of k[1]x...xk[m]? For example, when the length of the rope is 8, we cut it into three pieces of length 2, 3, and 3, and the maximum product obtained at this time is 18.

Title link : Cut the rope

code

/**
 * 标题：剪绳子
 * 题目描述
 * 给你一根长度为n的绳子，请把绳子剪成整数长的m段（m、n都是整数，n>1并且m>1，m<=n），每段绳子的长度记为k[1],...,k[m]。
 * 请问k[1]x...xk[m]可能的最大乘积是多少？例如，当绳子的长度是8时，我们把它剪成长度分别为2、3、3的三段，此时得到的最大乘积是18。
 * 题目链接：
 * https://www.nowcoder.com/practice/57d85990ba5b440ab888fc72b0751bf8?tpId=13&&tqId=33257&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
 */
public class Jz67 {
    /**
     * 方法一：贪心
     * 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。
     * <p>
     * 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。
     *
     * @param target
     * @return
     */
    public int cutRope(int target) {
        if (target < 2) {
            return 0;
        }
        if (target == 2) {
            return 1;
        }
        if (target == 3) {
            return 2;
        }
        int timesOf3 = target / 3;
        if (target - timesOf3 * 3 == 1) {
            timesOf3--;
        }
        int timesOf2 = (target - timesOf3 * 3) / 2;
        return ((int) Math.pow(3, timesOf3)) * ((int) Math.pow(2, timesOf2));
    }

    /**
     * 方法二：动态规划
     *
     * @param target
     * @return
     */
    public int cutRope1(int target) {
        int[] dp = new int[target + 1];
        dp[1] = 1;
        for (int i = 2; i <= target; 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[target];
    }

    public static void main(String[] args) {
        Jz67 jz67 = new Jz67();
        System.out.println(jz67.cutRope(2));
        System.out.println(jz67.cutRope(8));
        System.out.println("动态规划");
        System.out.println(jz67.cutRope1(2));
        System.out.println(jz67.cutRope1(8));
    }
}

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