Question: you a string of length n, please cut the rope segment length integer m (m, n are integers, n> 1 and m> 1), the length of each rope is referred to as k [0] , k [1], ..., k [m]. Will k [0] xk [1] x ... xk [m] maximum possible product is how much? For example, when the length of the rope is 8:00, we cut it into lengths of three sections 2,3,3 of the product obtained at this time is the maximum 18.
Input: length n
Output: maximum product rope segment
Ideas:
Dynamic Programming
Defined function f (n) is the maximum segment length to the length of the product is cut into segments n rope
, sub-optimal solution of the original problem of optimal solutions may be composed.
Between the sub-sub-problem and the problem of overlapping, and therefore the bottom-up method (open array storing sub-optimal solution of the problem) is calculated from f (2), f (3), f (4) ... f (n), the output of f (n).
Special case: Because the number of segments is greater than 1, so the length of the rope when the result is 2, 3, respectively 1 and 2.
greedy
Greedy strategy:
When n> = 5, as the length of the rope 3 is cut, when the remaining length of the rope 4, is cut into two lengths of rope 2.
Prove optimal solution:
When n> = 5, 3 * ( n-3)> n, 2 * (n-2)> n and 3 * (n-3)> = 2 * (n-2)
When the length of rope is greater than the rest of the It is equal to 5, as cut length of 3.
When n is 4, cut length of cord 2 causes the product of the maximum section 2.
Code:
Dynamic Programming
class Solution {
public:
int cutRope(int number) {
if(number<2)
return 0;
if(number==2)
return 1;
if(number==3)
return 2;
int* result=new int[number+1];
result[0]=0;
result[1]=1;
result[2]=2;
result[3]=3;
int max=0;
for(int i=4;i<=number;++i)
{
for(int j=1;j<=i/2;++j)
{
int temp=result[j]*result[i-j];
if(temp>max)
max=temp;
}
result[i]=max;
}
max=result[number];
delete[] result;
return max;
}
};greedy
int cutRope(int number) {
if(number<2)
return 0;
if(number==2)
return 1;
if(number==3)
return 2;
int result;
int timesOf3=number/3;
if(number-3*timesOf3==1)
{
--timesOf3;
return pow(3,timesOf3)*4;
}
return pow(3,timesOf3)*(number-3*timesOf3);
}Complexity analysis:
Dynamic programming algorithm: time complexity , space complexity O (n)
Greedy algorithm: time complexity of O (1), the space complexity of O (1)
