剑指offer--剪绳子

题目描述：给你一根长度为n的绳子，请把绳子剪成m段（m，n均为正数且均大于1）每段绳子的长度之间乘积，最大乘积为多少？
思路：动态规划

//剪绳子，求最大乘机 8=2+3+3（2*3*3=18)
package Function;
//products[0]=0,products[1]=1,products[2]=2,products[3]=3,
//我的理解是，当length=4时，那么products[4]=products[1]*products[3]或者products[4]=products[2]*products[2]。
//动态规划
public class CutRope14 {
//绳子长度
public int cutRope(int length){
if (length < 2)
return 0;
if (length == 2)
return 1;
if (length == 3)
return 2;
int[] products = new int[length + 1];
products[0] = 0;
products[1] = 1;
products[2] = 2;
products[3] = 3;
int max = 0;
//products[4]=products[1]*products[3]
// 或者products[4]=products[2]*products[2]
for (int i = 4; i <= length; i++) {
for (int j = 1; j <= i/2; j++) {
int product = products[j]*products[i-j];
if (max < product)
max = product;
products[i] = max;
}
}
max = products[length];
return max;
}
public static void main(String[] args) {
CutRope14 rope = new CutRope14();
System.out.println(rope.cutRope(8));
}

}

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