[编程珠玑]aha,algorithms

Rotate a one-dimensional vector of n elements left by i positions.

Solution1

 1 void Solution(char* arr,int n,int rotdist)
 2 {
 3     int i;
 4     //循环n和rotdist的最大公约数次
 5     for(i=0;i<gcd(n,rotdist);++i)
 6     {
 7         char t=arr[i];
 8         int index=i;
 9         while((index+rotdist)%n!=i)
10         {
11             arr[index]=arr[(index+rotdist)%n];
12             index=(index+rotdist)%n;
13         }
14         arr[index]=t;
15     }
16 }

 

Solution2

View Code

 1 //swapthe two segments of the vector ab to be the vector ba,
 2 //where a represents the first i elements of x.Suppose a is shorter than b.
 3 //Divide b into bl and br so that br is the same length as a.
 4 //Swap a and br to transform ablbr into brbla.The sequence a is 
 5 //in its final place,so we can focus on swapping the two parts of b.
 6 char* arr;
 7 
 8 void swap(int s1,int e1,int s2,int e2)
 9 {
10     int i,j;
11     for(i=s1,j=s2;i<=e1;++i,++j)
12     {
13         arr[i]=arr[j];
14     }
15 }
16 
17 void Solution(char* arr,int start,int end,int rotdist)
18 {
19     int n=end-start+1;
20     if(rotdist==n-rotdist)
21     {
22         swap(start,start+rotdist-1,start+rotdist,end);
23         return;
24     }
25     else if(rotdist<n-rotdist)//左边小
26     {
27         swap(start,start+rotdist-1,end-rotdist+1,end);
28         Solution(arr,start,end-rotdist,rotdist);
29     }
30     else
31     {
32         swap(start,end-rotdist,start+rotdist,end);
33         Solution(arr,end-rotdist+1,end,2*rotdist+start-end-1);
34     }
35 }

 Solution3(Best one)

View Code

1 reverse(0,rotdist-1);
2 reverse(rotdist,n-1);
3 reverse(0,n-1);

 

PS:Solve great common divisor

View Code

 1 int gcd(int i,int j)
 2 {
 3      while(i!=j)
 4      {
 5            if(i>j)
 6              i=i-j;
 7            else
 8              j=j-i;
 9      }
10      return i;
11 }